代码面试最常用的10大算法

我不是女神ヾ 2022-08-27 02:25 167阅读 0赞

在程序员的职业生涯中，算法亦算是一门基础课程，尤其是在面试的时候，很多公司都会让程序员编写一些算法实例，例如快速排序、二叉树查找等等。  
  
本文总结了程序员在代码面试中最常遇到的10大算法类型，想要真正了解这些算法的原理，还需程序员们花些功夫。  
  
**1.String/Array/Matrix**  
  
在Java中，String是一个包含char数组和其它字段、方法的类。如果没有IDE自动完成代码，下面这个方法大家应该记住：

toCharArray() //get char array of a String
    Arrays.sort()  //sort an array
    Arrays.toString(char[] a) //convert to string
    charAt(int x) //get a char at the specific index
    length() //string length
    length //array size 
    substring(int beginIndex) 
    substring(int beginIndex, int endIndex)
    Integer.valueOf()//string to integer
    String.valueOf()/integer to string

String/arrays很容易理解，但与它们有关的问题常常需要高级的算法去解决，例如动态编程、递归等。  
下面列出一些需要高级算法才能解决的经典问题：  
  
[Evaluate Reverse Polish Notation][]  
[Longest Palindromic Substring][]  
[单词分割][Link 1]  
[字梯][Link 2]  
[Median of Two Sorted Arrays][]  
[正则表达式匹配][Link 3]  
[合并间隔][Link 4]  
[插入间隔][Link 5]  
[Two Sum][]  
[3Sum][]  
[4Sum][]  
[3Sum Closest][]  
[String to Integer][]  
[合并排序数组][Link 6]  
[Valid Parentheses][]  
[实现strStr()][strStr]  
[Set Matrix Zeroes][]  
[搜索插入位置][Link 7]  
[Longest Consecutive Sequence][]  
[Valid Palindrome][Valid Parentheses]  
[螺旋矩阵][Link 8]  
[搜索一个二维矩阵][Link 9]  
[旋转图像][Link 10]  
[三角形][Link 11]  
[Distinct Subsequences Total][]  
[Maximum Subarray][]  
[删除重复的排序数组][Link 12]  
[删除重复的排序数组2][2]  
[查找没有重复的最长子串][Link 13]  
[包含两个独特字符的最长子串][Link 14]  
[Palindrome Partitioning][]  
  
**2.链表**  
  
在Java中实现链表是非常简单的，每个节点都有一个值，然后把它链接到下一个节点。

Java code

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      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>int</code>&nbsp; 
      <code>val;</code> 
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      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>Node&nbsp;next;</code> 
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      <code>&nbsp;</code>&nbsp; 
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      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>Node(</code> 
      <code>int</code>&nbsp; 
      <code>x)&nbsp;{ </code> 
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      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>val&nbsp;=&nbsp;x;</code> 
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      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>next&nbsp;=&nbsp;</code> 
      <code>null</code> 
      <code>;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>}</code> 
     </div> 
    </div> </td> 
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比较流行的两个链表例子就是栈和队列。  
  
栈（Stack）

Java code

[?][Link 15]

队列（Queue）

Java code

[?][Link 15]

值得一提的是，Java标准库中已经包含一个叫做Stack的类，链表也可以作为一个队列使用（add()和remove()）。（链表实现队列接口）如果你在面试过程中，需要用到栈或队列解决问题时，你可以直接使用它们。  
  
在实际中，需要用到链表的算法有：  
  
[插入两个数字][Link 16]  
[重新排序列表][Link 17]  
[链表周期][Link 18]  
[Copy List with Random Pointer][]  
[合并两个有序列表][Link 19]  
[合并多个排序列表][Link 20]  
[从排序列表中删除重复的][Link 21]  
[分区列表][Link 22]  
[LRU缓存][LRU]  
  
**3.树&堆**  
  
这里的树通常是指二叉树。

Java code

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    <div> 
     <div> 
      <code>class</code>&nbsp; 
      <code>TreeNode{ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>int</code>&nbsp; 
      <code>value;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>TreeNode&nbsp;left;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>TreeNode&nbsp;right;</code> 
     </div> 
     <div> 
      <code>}&nbsp;</code> 
     </div> 
    </div> </td> 
  </tr> 
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</table>

下面是一些与二叉树有关的概念：  
**二叉树搜索：**对于所有节点，顺序是：left children <= current node <= right children；  
**平衡vs.非平衡：**它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1，并且左右两个子树都是一棵平衡二叉树；  
**满二叉树：**除最后一层无任何子节点外，每一层上的所有结点都有两个子结点；  
**完美二叉树（Perfect Binary Tree）：**一个满二叉树，所有叶子都在同一个深度或同一级，并且每个父节点都有两个子节点；  
**完全二叉树：**若设二叉树的深度为h，除第 h 层外，其它各层 (1～h-1) 的结点数都达到最大个数，第 h 层所有的结点都连续集中在最左边，这就是完全二叉树。  
  
堆（Heap）是一个基于树的数据结构，也可以称为优先队列（ PriorityQueue），在队列中，调度程序反复提取队列中第一个作业并运行，因而实际情况中某些时间较短的任务将等待很长时间才能结束，或者某些不短小，但具有重要性的作业，同样应当具有优先权。堆即为解决此类问题设计的一种数据结构。  
  
下面列出一些基于二叉树和堆的算法：  
  
[二叉树前序遍历][Link 23]  
[二叉树中序遍历][Link 24]  
[二叉树后序遍历][Link 25]  
[字梯][Link 2]  
[验证二叉查找树][Link 26]  
[把二叉树变平放到链表里][Link 27]  
[二叉树路径和][Link 28]  
[从前序和后序构建二叉树][Link 29]  
[把有序数组转换为二叉查找树][Link 30]  
[把有序列表转为二叉查找树][Link 31]  
[最小深度二叉树][Link 32]  
[二叉树最大路径和][Link 33]  
[平衡二叉树][Link 34]  
  
**4.Graph**  
  
与Graph相关的问题主要集中在深度优先搜索和宽度优先搜索。深度优先搜索非常简单，你可以从根节点开始循环整个邻居节点。下面是一个非常简单的宽度优先搜索例子，核心是用队列去存储节点。  
  
![1398320596_556431.jpg][]  
  
第一步，定义一个GraphNode

Java code

[?][Link 15]

第二步，定义一个队列

Java code

[?][Link 15]

第三步，使用队列进行宽度优先搜索

Java code

[?][Link 15]

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      <code>public</code>&nbsp; 
      <code>class</code>&nbsp; 
      <code>GraphTest&nbsp;{ </code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>public</code>&nbsp; 
      <code>static</code>&nbsp; 
      <code>void</code>&nbsp; 
      <code>main(String[]&nbsp;args)&nbsp;{ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>GraphNode&nbsp;n1&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode(</code> 
      <code>1</code> 
      <code>);&nbsp;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>GraphNode&nbsp;n2&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode(</code> 
      <code>2</code> 
      <code>);&nbsp;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>GraphNode&nbsp;n3&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode(</code> 
      <code>3</code> 
      <code>);&nbsp;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>GraphNode&nbsp;n4&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode(</code> 
      <code>4</code> 
      <code>);&nbsp;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>GraphNode&nbsp;n5&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode(</code> 
      <code>5</code> 
      <code>);&nbsp;</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>n1.neighbors&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode[]{n2,n3,n5};</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>n2.neighbors&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode[]{n1,n4};</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>n3.neighbors&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode[]{n1,n4,n5};</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>n4.neighbors&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode[]{n2,n3,n5};</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>n5.neighbors&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>GraphNode[]{n1,n3,n4};</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>breathFirstSearch(n1,&nbsp;</code> 
      <code>5</code> 
      <code>);</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>public</code>&nbsp; 
      <code>static</code>&nbsp; 
      <code>void</code>&nbsp; 
      <code>breathFirstSearch(GraphNode&nbsp;root,&nbsp;</code> 
      <code>int</code>&nbsp; 
      <code>x){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>if</code> 
      <code>(root.val&nbsp;==&nbsp;x)</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>System.out.println(</code> 
      <code>"find&nbsp;in&nbsp;root"</code> 
      <code>);</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>Queue&nbsp;queue&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>Queue();</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>root.visited&nbsp;=&nbsp;</code> 
      <code>true</code> 
      <code>;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>queue.enqueue(root);</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>while</code> 
      <code>(queue.first&nbsp;!=&nbsp;</code> 
      <code>null</code> 
      <code>){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>GraphNode&nbsp;c&nbsp;=&nbsp;(GraphNode)&nbsp;queue.dequeue();</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>for</code> 
      <code>(GraphNode&nbsp;n:&nbsp;c.neighbors){ </code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>if</code> 
      <code>(!n.visited){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>System.out.print(n&nbsp;+&nbsp;</code> 
      <code>"&nbsp;"</code> 
      <code>);</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>n.visited&nbsp;=&nbsp;</code> 
      <code>true</code> 
      <code>;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>if</code> 
      <code>(n.val&nbsp;==&nbsp;x)</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>System.out.println(</code> 
      <code>"Find&nbsp;"</code> 
      <code>+n);</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>queue.enqueue(n);</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>}</code> 
     </div> 
    </div> </td> 
  </tr> 
 </tbody> 
</table>

输出结果：  
value: 2 value: 3 value: 5 Find value: 5  
value: 4  
  
实际中，基于Graph需要经常用到的算法：  
  
[克隆Graph][Graph]

**5.排序**  
  
不同排序算法的时间复杂度，大家可以到wiki上查看它们的基本思想。  
  
![1398337090_137253.jpg][]  
  
BinSort、Radix Sort和CountSort使用了不同的假设，所有，它们不是一般的排序方法。  
下面是这些算法的具体实例，另外，你还可以阅读：[Java开发者在实际操作中是如何排序的][Java]。  
  
[归并排序][Link 35]  
[快速排序][Link 36]  
[插入排序][Link 37]  
  
**6.递归和迭代**  
  
下面通过一个例子来说明什么是递归。  
  
**问题：**  
  
这里有n个台阶，每次能爬1或2节，请问有多少种爬法？  
  
步骤1：查找n和n-1之间的关系  
  
为了获得n，这里有两种方法：一个是从第一节台阶到n-1或者从2到n-2。如果f(n)种爬法刚好是爬到n节，那么f(n)=f(n-1)+f(n-2)。  
  
步骤2：确保开始条件是正确的  
  
f(0) = 0;  
f(1) = 1;

Java code

[?][Link 15]

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      <code>public</code>&nbsp; 
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      <code>int</code>&nbsp; 
      <code>f(</code> 
      <code>int</code>&nbsp; 
      <code>n){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>if</code> 
      <code>(n&nbsp;&lt;=&nbsp;</code> 
      <code>2</code> 
      <code>)&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>n;</code> 
     </div> 
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      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>int</code>&nbsp; 
      <code>x&nbsp;=&nbsp;f(n-</code> 
      <code>1</code> 
      <code>)&nbsp;+&nbsp;f(n-</code> 
      <code>2</code> 
      <code>);</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>x;</code> 
     </div> 
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      <code>}</code> 
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递归方法的时间复杂度指数为n，这里会有很多冗余计算。

Java code

[?][Link 15]

该递归可以很简单地转换为迭代。

Java code

[?][Link 15]

在这个例子中，迭代花费的时间要少些。关于迭代和递归，你可以去 [这里][Link 38]看看。  
  
**7.动态规划**  
  
动态规划主要用来解决如下技术问题：  
  
通过较小的子例来解决一个实例；  
对于一个较小的实例，可能需要许多个解决方案；  
把较小实例的解决方案存储在一个表中，一旦遇上，就很容易解决；  
附加空间用来节省时间。  
上面所列的爬台阶问题完全符合这四个属性，因此，可以使用动态规划来解决：

Java code

[?][Link 15]

<table> 
 <tbody> 
  <tr> 
   <td> 
    <div>
      1 
    </div> 
    <div>
      2 
    </div> 
    <div>
      3 
    </div> 
    <div>
      4 
    </div> 
    <div>
      5 
    </div> 
    <div>
      6 
    </div> 
    <div>
      7 
    </div> 
    <div>
      8 
    </div> 
    <div>
      9 
    </div> 
    <div>
      10 
    </div> 
    <div>
      11 
    </div> 
    <div>
      12 
    </div> </td> 
   <td> 
    <div> 
     <div> 
      <code>public</code>&nbsp; 
      <code>static</code>&nbsp; 
      <code>int</code> 
      <code>[]&nbsp;A&nbsp;=&nbsp;</code> 
      <code>new</code>&nbsp; 
      <code>int</code> 
      <code>[</code> 
      <code>100</code> 
      <code>];</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>public</code>&nbsp; 
      <code>static</code>&nbsp; 
      <code>int</code>&nbsp; 
      <code>f3(</code> 
      <code>int</code>&nbsp; 
      <code>n)&nbsp;{ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>if</code>&nbsp; 
      <code>(n&nbsp;&lt;=&nbsp;</code> 
      <code>2</code> 
      <code>)</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>A[n]=&nbsp;n;</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>if</code> 
      <code>(A[n]&nbsp;&gt;&nbsp;</code> 
      <code>0</code> 
      <code>)</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>A[n];</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>else</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>A[n]&nbsp;=&nbsp;f3(n-</code> 
      <code>1</code> 
      <code>)&nbsp;+&nbsp;f3(n-</code> 
      <code>2</code> 
      <code>);</code> 
      <code>//store&nbsp;results&nbsp;so&nbsp;only&nbsp;calculate&nbsp;once!</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>A[n];</code> 
     </div> 
     <div> 
      <code>}</code> 
     </div> 
    </div> </td> 
  </tr> 
 </tbody> 
</table>

一些基于动态规划的算法：  
[编辑距离][Link 39]  
[最长回文子串][Longest Palindromic Substring]  
[单词分割][Link 1]  
[最大的子数组][Maximum Subarray]  
  
**8.位操作**  
  
位操作符：  
![1398337471_969269.jpg][]  
  
从一个给定的数n中找位i（i从0开始，然后向右开始）

Java code

[?][Link 15]

<table> 
 <tbody> 
  <tr> 
   <td> 
    <div>
      1 
    </div> 
    <div>
      2 
    </div> 
    <div>
      3 
    </div> 
    <div>
      4 
    </div> 
    <div>
      5 
    </div> 
    <div>
      6 
    </div> 
    <div>
      7 
    </div> 
    <div>
      8 
    </div> 
    <div>
      9 
    </div> </td> 
   <td> 
    <div> 
     <div> 
      <code>public</code>&nbsp; 
      <code>static</code>&nbsp; 
      <code>boolean</code>&nbsp; 
      <code>getBit(</code> 
      <code>int</code>&nbsp; 
      <code>num,&nbsp;</code> 
      <code>int</code>&nbsp; 
      <code>i){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>int</code>&nbsp; 
      <code>result&nbsp;=&nbsp;num&nbsp;&amp;&nbsp;(</code> 
      <code>1</code> 
      <code>&lt;&lt;i);</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>if</code> 
      <code>(result&nbsp;==&nbsp;</code> 
      <code>0</code> 
      <code>){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>false</code> 
      <code>;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
      <code>else</code> 
      <code>{ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>true</code> 
      <code>;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>}</code> 
     </div> 
    </div> </td> 
  </tr> 
 </tbody> 
</table>

例如，获取10的第二位：

Java code

[?][Link 15]

<table> 
 <tbody> 
  <tr> 
   <td> 
    <div>
      1 
    </div> 
    <div>
      2 
    </div> 
    <div>
      3 
    </div> 
    <div>
      4 
    </div> </td> 
   <td> 
    <div> 
     <div> 
      <code>i=</code> 
      <code>1</code> 
      <code>,&nbsp;n=</code> 
      <code>10</code> 
     </div> 
     <div> 
      <code>1</code> 
      <code>&lt;&lt;</code> 
      <code>1</code> 
      <code>=&nbsp;</code> 
      <code>10</code> 
     </div> 
     <div> 
      <code>1010</code> 
      <code>&amp;</code> 
      <code>10</code> 
      <code>=</code> 
      <code>10</code> 
     </div> 
     <div> 
      <code>10</code>&nbsp; 
      <code>is&nbsp;not&nbsp;</code> 
      <code>0</code> 
      <code>,&nbsp;so&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>true</code> 
      <code>;</code> 
     </div> 
    </div> </td> 
  </tr> 
 </tbody> 
</table>

典型的位算法：  
[Find Single Number][]  
[Maximum Binary Gap][]  
  
**9.概率**  
  
通常要解决概率相关问题，都需要很好地格式化问题，下面提供一个简单的例子：  
有50个人在一个房间，那么有两个人是同一天生日的可能性有多大？（忽略闰年，即一年有365天）  
  
算法：

Java code

[?][Link 15]

<table> 
 <tbody> 
  <tr> 
   <td> 
    <div>
      1 
    </div> 
    <div>
      2 
    </div> 
    <div>
      3 
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    <div>
      4 
    </div> 
    <div>
      5 
    </div> 
    <div>
      6 
    </div> 
    <div>
      7 
    </div> 
    <div>
      8 
    </div> 
    <div>
      9 
    </div> 
    <div>
      10 
    </div> </td> 
   <td> 
    <div> 
     <div> 
      <code>public</code>&nbsp; 
      <code>static</code>&nbsp; 
      <code>double</code>&nbsp; 
      <code>caculateProbability(</code> 
      <code>int</code>&nbsp; 
      <code>n){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>double</code>&nbsp; 
      <code>x&nbsp;=&nbsp;</code> 
      <code>1</code> 
      <code>;&nbsp;</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>for</code> 
      <code>(</code> 
      <code>int</code>&nbsp; 
      <code>i=</code> 
      <code>0</code> 
      <code>;&nbsp;i&lt;n;&nbsp;i++){ </code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>x&nbsp;*=&nbsp;&nbsp;(</code> 
      <code>365.0</code> 
      <code>-i)/</code> 
      <code>365.0</code> 
      <code>;</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>}</code> 
     </div> 
     <div> 
      <code>&nbsp;</code>&nbsp; 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>double</code>&nbsp; 
      <code>pro&nbsp;=&nbsp;Math.round((</code> 
      <code>1</code> 
      <code>-x)&nbsp;*&nbsp;</code> 
      <code>100</code> 
      <code>);</code> 
     </div> 
     <div> 
      <code>&nbsp;&nbsp;&nbsp;&nbsp;</code> 
      <code>return</code>&nbsp; 
      <code>pro/</code> 
      <code>100</code> 
      <code>;</code> 
     </div> 
     <div> 
      <code>}</code> 
     </div> 
    </div> </td> 
  </tr> 
 </tbody> 
</table>

结果：calculateProbability(50) = 0.97  
  
**10.组合和排列**  
  
组合和排列的主要差别在于顺序是否重要。  
  
例1：  
  
1、2、3、4、5这5个数字，输出不同的顺序，其中4不可以排在第三位，3和5不能相邻，请问有多少种组合？  
  
例2：  
  
有5个香蕉、4个梨、3个苹果，假设每种水果都是一样的，请问有多少种不同的组合？  
  
基于它们的一些常见算法  
[排列][Link 40]  
[排列2][2 1]  
[排列顺序][Link 41]  
  
**来自：** [ProgramCreek][]

转自：[http://bbs.csdn.net/topics/390768965][http_bbs.csdn.net_topics_390768965]

[Evaluate Reverse Polish Notation]: http://www.programcreek.com/2012/12/leetcode-evaluate-reverse-polish-notation/
[Longest Palindromic Substring]: http://www.programcreek.com/2013/12/leetcode-solution-of-longest-palindromic-substring-java/
[Link 1]: http://www.programcreek.com/2012/12/leetcode-solution-word-break/
[Link 2]: http://www.programcreek.com/2012/12/leetcode-word-ladder/
[Median of Two Sorted Arrays]: http://www.programcreek.com/2012/12/leetcode-median-of-two-sorted-arrays-java/
[Link 3]: http://www.programcreek.com/2012/12/leetcode-regular-expression-matching-in-java/
[Link 4]: http://merge%20intervals/
[Link 5]: http://www.programcreek.com/2012/12/leetcode-insert-interval/
[Two Sum]: http://www.programcreek.com/2012/12/leetcode-solution-of-two-sum-in-java/
[3Sum]: http://www.programcreek.com/2012/12/leetcode-3sum/
[4Sum]: http://www.programcreek.com/2013/02/leetcode-4sum-java/
[3Sum Closest]: http://www.programcreek.com/2013/02/leetcode-3sum-closest-java/
[String to Integer]: http://www.programcreek.com/2012/12/leetcode-string-to-integer-atoi/
[Link 6]: http://www.programcreek.com/2012/12/leetcode-merge-sorted-array-java/
[Valid Parentheses]: http://www.programcreek.com/2013/01/leetcode-valid-palindrome-java/
[strStr]: http://www.programcreek.com/2012/12/leetcode-implement-strstr-java/
[Set Matrix Zeroes]: http://www.programcreek.com/2012/12/leetcode-set-matrix-zeroes-java/
[Link 7]: http://www.programcreek.com/2013/01/leetcode-search-insert-position/
[Longest Consecutive Sequence]: http://www.programcreek.com/2013/01/leetcode-longest-consecutive-sequence-java/
[Link 8]: http://www.programcreek.com/2013/01/leetcode-spiral-matrix-java/
[Link 9]: http://www.programcreek.com/2013/01/leetcode-search-a-2d-matrix-java/
[Link 10]: http://www.programcreek.com/2013/01/leetcode-rotate-image-java/
[Link 11]: http://www.programcreek.com/2013/01/leetcode-triangle-java/
[Distinct Subsequences Total]: http://www.programcreek.com/2013/01/leetcode-distinct-subsequences-total-java/
[Maximum Subarray]: http://www.programcreek.com/2013/02/leetcode-maximum-subarray-java/
[Link 12]: http://www.programcreek.com/2013/01/leetcode-remove-duplicates-from-sorted-array-java/
[2]: http://www.programcreek.com/2013/01/leetcode-remove-duplicates-from-sorted-array-ii-java/
[Link 13]: http://www.programcreek.com/2013/02/leetcode-longest-substring-without-repeating-characters-java/
[Link 14]: http://www.programcreek.com/2013/02/longest-substring-which-contains-2-unique-characters/
[Palindrome Partitioning]: http://www.programcreek.com/2013/03/leetcode-palindrome-partitioning-java/
[Link 15]: http://bbs.csdn.net/topics/390768965#clipboardWindow
[Link 16]: http://www.programcreek.com/2012/12/add-two-numbers/
[Link 17]: http://www.programcreek.com/2013/12/in-place-reorder-a-singly-linked-list-in-java/
[Link 18]: http://www.programcreek.com/2012/12/leetcode-linked-list-cycle/
[Copy List with Random Pointer]: http://www.programcreek.com/2012/12/leetcode-copy-list-with-random-pointer/
[Link 19]: http://www.programcreek.com/2012/12/leetcode-merge-two-sorted-lists-java/
[Link 20]: http://www.programcreek.com/2013/02/leetcode-merge-k-sorted-lists-java/
[Link 21]: http://www.programcreek.com/2013/01/leetcode-remove-duplicates-from-sorted-list/
[Link 22]: http://www.programcreek.com/2013/02/leetcode-partition-list-java/
[LRU]: http://www.programcreek.com/2013/03/leetcode-lru-cache-java/
[Link 23]: http://www.programcreek.com/2012/12/leetcode-solution-for-binary-tree-preorder-traversal-in-java/
[Link 24]: http://www.programcreek.com/2012/12/leetcode-solution-of-binary-tree-inorder-traversal-in-java/
[Link 25]: http://www.programcreek.com/2012/12/leetcode-solution-of-iterative-binary-tree-postorder-traversal-in-java/
[Link 26]: http://www.programcreek.com/2012/12/leetcode-validate-binary-search-tree-java/
[Link 27]: http://www.programcreek.com/2013/01/leetcode-flatten-binary-tree-to-linked-list/
[Link 28]: http://www.programcreek.com/2013/01/leetcode-path-sum/
[Link 29]: http://www.programcreek.com/2013/01/construct-binary-tree-from-inorder-and-postorder-traversal/
[Link 30]: http://www.programcreek.com/2013/01/leetcode-convert-sorted-array-to-binary-search-tree-java/
[Link 31]: http://www.programcreek.com/2013/01/leetcode-convert-sorted-list-to-binary-search-tree-java/
[Link 32]: http://www.programcreek.com/2013/02/leetcode-minimum-depth-of-binary-tree-java/
[Link 33]: http://www.programcreek.com/2013/02/leetcode-binary-tree-maximum-path-sum-java/
[Link 34]: http://www.programcreek.com/2013/02/leetcode-balanced-binary-tree-java/
[1398320596_556431.jpg]: http://img.bbs.csdn.net/upload/201404/24/1398320596_556431.jpg
[Graph]: http://www.programcreek.com/2012/12/leetcode-clone-graph-java/
[1398337090_137253.jpg]: http://img.bbs.csdn.net/upload/201404/24/1398337090_137253.jpg
[Java]: http://www.programcreek.com/2014/03/how-developers-sort-in-java/
[Link 35]: http://www.programcreek.com/2012/11/leetcode-solution-merge-sort-linkedlist-in-java/
[Link 36]: http://www.programcreek.com/2012/11/quicksort-array-in-java/
[Link 37]: http://www.programcreek.com/2012/11/leetcode-solution-sort-a-linked-list-using-insertion-sort-in-java/
[Link 38]: http://www.programcreek.com/2012/10/iteration-vs-recursion-in-java/
[Link 39]: http://www.programcreek.com/2013/12/edit-distance-in-java/
[1398337471_969269.jpg]: http://img.bbs.csdn.net/upload/201404/24/1398337471_969269.jpg
[Find Single Number]: http://www.programcreek.com/2012/12/leetcode-solution-of-single-number-in-java/
[Maximum Binary Gap]: http://www.programcreek.com/2013/02/twitter-codility-problem-max-binary-gap/
[Link 40]: http://www.programcreek.com/2013/02/leetcode-permutations-java/
[2 1]: http://www.programcreek.com/2013/02/leetcode-permutations-ii-java/
[Link 41]: http://www.programcreek.com/2013/02/leetcode-permutation-sequence-java/
[ProgramCreek]: http://www.programcreek.com/2012/11/top-10-algorithms-for-coding-interview/
[http_bbs.csdn.net_topics_390768965]: http://bbs.csdn.net/topics/390768965