ARTS LeetCode1 TwoSum

傷城~ 2022-03-01 21:16 194阅读 0赞

two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = \[2, 7, 11, 15\], target = 9,

Because nums\[0\] + nums\[1\] = 2 + 7 = 9,  
return \[0, 1\].

两个数求和

给定一个整数数组，返回两个数字的索引，使它们相加到特定目标值。

您可以假设每个输入只有一个解决方案，并且您可能不会两次使用相同的元素。

我的思路是：数组中任意两个元素之和等于目标值，并且返回这两个元素的下标。遍历两次数组，第一次取的值当做第一个元素，然后用目标值相减得到另外一个值，然后判断第二个值是否存在于数组中，如果存在就返回两个元素的下标，如果不存在，则返回null。

code:

class Solution {
        public int[] twoSum(int[] nums, int target) {
               int[] result = new int[2];
            for(int i=0;i<nums.length;i++){
                int other = target - nums[i] ;
                for(int j=i+1;j<nums.length;j++){
                    if(other  == nums[j]){
                        result[0] = i;
                        result[1] = j;
                    }
                }
            }
            return  result;
        }
    }

结果分析：

Runtime: 23 ms, faster than 32.78% of Java online submissions for Two Sum.
    Memory Usage: 38.6 MB, less than 39.16% of Java online submissions for Two Sum.

code2:

class Solution {
        public int[] twoSum(int[] nums, int target) {
               
            for(int i=0;i<nums.length;i++){
                for(int j=i+1;j<nums.length;j++){
                    if(target - nums[i]  == nums[j]){
                        return new int[]{i,j};
                    }
                }
            }
            return  null;
        }
    }

结果分析：

Runtime: 15 ms, faster than 44.10% of Java online submissions for Two Sum.
    Memory Usage: 38.4 MB, less than 50.25% of Java online submissions for Two Sum.

上述解法在时间复杂度是O(n),在空间复杂度是O(1)。

下面是别人的解法，利用空间换时间的想法，来实现：

利用hashtable来解决，下面的时间复杂度是O(n)，空间复杂度也是O(n)，比上面的两种做法效率高了很多，当时自己第一反应就是用数据，并没有想到用hashtable来处理，所以选择合适的数据结构，可以提高程序的运行效率。

class Solution {
        public int[] twoSum(int[] nums, int target) {
            Map<Integer,Integer> map = new HashMap<Integer,Integer>();
           for(int i=0;i<nums.length;i++){
               map.put(nums[i],i);
           }
           for(int i=0;i<nums.length;i++){
               int result = target - nums[i];
               if(map.containsKey(result) && map.get(result) !=i){
                   return new int[]{i,map.get(result)};
               }
           }
          throw new IllegalArgumentException("no two sum solution");
        }
    }

Runtime: 2 ms, faster than 99.93% of Java online submissions for Two Sum.
    Memory Usage: 39.2 MB, less than 23.52% of Java online submissions for Two Sum.

在一次循环中做处理：

class Solution {
        public int[] twoSum(int[] nums, int target) {
            Map<Integer,Integer> map = new HashMap<Integer,Integer>();
           for(int i=0;i<nums.length;i++){
               int result = target - nums[i];
               if(map.containsKey(result) && map.get(result) !=i){
                   return new int[]{i,map.get(result)};
               }
               map.put(nums[i],i);
           }
          throw new IllegalArgumentException("no two sum solution");
        }
    }

Runtime: 2 ms, faster than 99.93% of Java online submissions for Two Sum.
    Memory Usage: 39.1 MB, less than 24.88% of Java online submissions for Two Sum.

看了讨论区又发现一种效率更高的算法，思路如下：  
1.先把原有数组copy到一个新数组中  
2.对新数组进行排序，让元素从小到大排列  
3.while循环找出两个num值  
4.循环原有数组找出值为第3步中找到的值的索引，此处分别讨论了num1和num2相等和不等的情况  
5.最后返回

class Solution {
        public int[] twoSum(int[] nums, int target) {
          int[] num2 = Arrays.copyOf(nums,nums.length);
            Arrays.sort(num2);
            int start = 0;
            int end = num2.length-1;
            int a = 0;int b = 0;
            while (start<end){
                int sum = num2[start] + num2[end];
                if(sum < target){
                    start++;
                }else if(sum > target){
                    end--;
                }else{
                    a = num2[start];
                    b = num2[end];
                    break;
                }
            }
            int[] index = new int[2];
            for(int i=0;i<nums.length;i++){
                if(nums[i] == a){
                    index[0] = i;
                    break;
                }
            }
            if(a!=b){
                for(int i=0;i<nums.length;i++){
                    if(nums[i] == b){
                        index[1] = i;
                        break;
                    }
                }
            }else{
                for(int i=0;i<nums.length;i++){
                    if(nums[i] == b && index[0] != i){
                        index[1] = i;
                        break;
                    }
                }
            }
            
            return  index;
        }
    }

Runtime: 2 ms, faster than 99.93% of Java online submissions for Two Sum.
    Memory Usage: 38.1 MB, less than 73.96% of Java online submissions for Two Sum.

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