[leetcode-排序]--75. Sort Colors

桃扇骨 2022-07-12 02:53 156阅读 0赞

Question 75. Sort Colors

> Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.
> 
> Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.
> 
> Note:  
> You are not suppose to use the library’s sort function for this problem.
> 
> Follow up:  
> A rather straight forward solution is a two-pass algorithm using counting sort.  
> First, iterate the array counting number of 0’s, 1’s, and 2’s, then overwrite array with total number of 0’s, then 1’s and followed by 2’s.
> 
> Could you come up with an one-pass algorithm using only constant space?
> 
> 中文：有一个数组由n个颜色对象组成，在、颜色对象分别是红、白、蓝。将他们排序，使得数组中颜色顺序是红、白、蓝。现在分别用0、1、2来表示红白蓝。 不得使用库函数的排序方法，
> 
> 对于两轮遍历来说比较简单，一轮遍历直接收集0和1和2的个数，然后第二轮遍历直接根据个数赋值，这是非常简单的做法。那么能不能通过一轮遍历实现呢？

解决思路：

# （1）两轮遍历解决 #

思路简单，第一轮收集0，1，2个数，第二轮赋值。

/** * 两轮遍历 * @param nums */
    public static void sortColors1(int[] nums) {
        int num0, num1, num2;
        num0 = num1 = num2 = 0;
    
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 0)
                num0++;
            else if (nums[i] == 1)
                num1++;
            else if (nums[i] == 2)
                num2++;
        }
    
        int i=0;
        while (num0>0){
            nums[i++]=0;
            num0--;
        }
        while (num1>0){
            nums[i++]=1;
            num1--;
        }
        while (num2>0){
            nums[i++]=2;
            num2--;
        }
    }

# (2)一轮遍历解决 #

一轮遍历: 使用了两个索引 :indexred 表示红色（0）的索；indexblue（ ） 表示蓝色的索引。这两个索引分别从首尾向中间夹逼;  
用i来从前往后遍历（ 注意，这里遍历的为边界是indexblue索引，而不是nums.lengtjh-1，只需要遍历到indexblue， indexblue后面的就全部是2了。此外遍历到nums.lengtjh-1可能会出现indexblue前面是1了而indexblue后面的2与前面的1交换，出现错误。 ）, 如果碰到0 就和indexred交换; 如果碰到2 就和indexblue交换。

实现源码：

public static void sortColors2(int[] nums) {
        int indexred = 0;
        int indexblue = nums.length-1;
        int temp;
    
        for (int i=0; i<=indexblue; i++){
            //只要nums[i]为2 就执行交换, 防止2,1,2 这种情况,这种情况下会不被排序
            while (nums[i] == 2 && i<indexblue){
                //将nums[i] 与 nums[indexblue] 交换
                temp = nums[i];
                nums[i] = nums[indexblue];
                nums[indexblue] = temp;
                indexblue--;
            }
    
            while(nums[i] == 0 && i>indexred){
                //将nums[i] 与 nums[indexred] 交换
                temp = nums[i];
                nums[i] = nums[indexred];
                nums[indexred] = temp;
    
                indexred++;
            }
        }
    }

本文的完整代码 github 地址：  
[https://github.com/leetcode-hust/leetcode/blob/master/louyuting/src/leetcode/Question75.java][https_github.com_leetcode-hust_leetcode_blob_master_louyuting_src_leetcode_Question75.java]

[https_github.com_leetcode-hust_leetcode_blob_master_louyuting_src_leetcode_Question75.java]: https://github.com/leetcode-hust/leetcode/blob/master/louyuting/src/leetcode/Question75.java