LeetCode Sort Colors

最新推荐文章于 2019-05-21 15:41:52 发布

weixin_30929295

最新推荐文章于 2019-05-21 15:41:52 发布

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CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/lailailai/p/3851896.html

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++;
            }
        }
    }
};