Given an array with n objects colored red, white or blue, sort them in-place 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.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

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 a one-pass algorithm using only constant space?

AC代码

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int l = 0, r = nums.size() - 1;
        for (int i = 0; i <= r; ++i) {
            while (nums[i] != 1) {
                if (i > r || i < l) break;
                else if (nums[i] == 0) swap(nums[i], nums[l++]);
                else if (nums[i] == 2) swap(nums[i], nums[r--]);
            }
        }
    }
};

他人代码

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

总结

一次遍历还要O（1）空间复杂度，用双指针，自己写的稍有复杂，别人写的更简单易懂。照抄自：https://www.cnblogs.com/grandyang/p/4341243.html