LeetCode 75. Sort Colors_psckage-优快云博客

本文介绍了一种不使用库排序函数的三色排序算法，通过单次遍历将数组中的0、1、2按顺序排列。文章详细解释了算法思路，并提供了实现代码。

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Description:
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?

本题如底下的follow up所言，最直观的做法是遍历一遍数组，得到0,1,2分别出现的次数，然后在原数组中按顺序重新填充0,1,2即可。不过本题要求只遍历一次数组，所以我想到了如下的做法：既然0和2是放在数组两边的，那么我就遍历一次该数组，每次遇到0就把他放到该数组的左端，遇到2就放到数组的右端，而1是在中间的，且在遍历数组的过程中，只要遇到了0或者2，就能把1替换到数组的中间部分，所以遇到1可以不用做处理，基于这种思想写下了如下代码：

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int len = nums.size();
        for(int i = 0, k = 0, j = len - 1; i < j && k <= j;) {
            if(nums[k] == 0) {
                int temp = nums[i];
                nums[i] = nums[k];
                nums[k] = temp;
                if(k == i || nums[k] == 1)
                    k++;
                i++;
            }
            else if(nums[k] == 2) {
                int temp = nums[j];
                nums[j] = nums[k];
                nums[k] = temp;
                if(k == j || nums[k] == 1)
                    k++;
                j--;
            }
            else {
                k++;
            }
        }
    }
};