LeetCode Sort Colors

class Solution {
public:
    void sortColors(int A[], int n) {
        int cnt[3] = {0};
        for (int i = 0; i < n; i++) {
            cnt[A[i]]++;
        }
        int start = 0;
        for (int i=0; i<3; i++) {
            if (i > 0) start += cnt[i-1];
            for (int j=start; j < start + cnt[i]; j++) {
                A[j] = i;
            }
        }
    }
};

简化版计数排序，two-pass就two-pass

第二轮：

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.

click to show follow up.

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?

 

从discuss上找了一下one-pass的解法，挺精妙的。

class Solution {
public:
    void sortColors(int A[], int n) {
        if (A == NULL || n < 1) return;
        int lo = 0, hi = n;
        int i = 0;
        while (i < hi) {
            if (A[i] < 1) {
                swap(A[i++], A[lo++]);
            } else if (A[i] > 1) {
                swap(A[i], A[--hi]);
            } else {
                i++;
            }
        }
    }
};

用lo，i，hi几个索引变量把数组分为4个部分：[0,lo-1], [lo, i-1], [i, hi-1], [hi, end]。使用i变量进行遍历，根据当前值的情况：

1. A[i] == 0,则将其和A[lo]交换（如上图所示），因为0和1的部分是相邻的，0的部分向右扩展了一个元素（该元素原来是1），而1的部分则仿佛右移了一个单元

2. A[i] == 1,直接i++，将其归入1的部分

3. A[i] == 2,将其和A[hi-1]交换，这样2的部分向左扩展了一个位置因而hi=hi-1