LeetCode Sort Colors

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?

题意：就是在数组A[]中，将0,1,2重新排序。

思路：计数排序或者三指针来做，一个指针指向放red的位置，一个指针指向blue的位置，然后就能依次放了，中间的没有处理的就都是白色的了

class Solution {
public:
    void sortColors(int A[], int n) {
        if (A == NULL || n < 2) return;
        int num0 = 0, num1 = 0, num2 = 0;
        for (int i = 0; i < n; i++) {
            if (A[i] == 0) num0++;
            else if (A[i] == 1) num1++;
            else num2++;
        }

        int i = 0;
        while (num0--) A[i++] = 0;
        while (num1--) A[i++] = 1;
        while (num2--) A[i++] = 2;
    }
};

class Solution {
public:
    void sortColors(int A[], int n) {
        if (A == NULL || n < 2) return;
        int red = 0, blue = n-1;
        for (int i = red; i <= blue; i++) {
            if (A[i] == 0) swap(A[red++], A[i]);
            else if (A[i] == 2) swap(A[blue--], A[i--]);
        }
    }
};