LEETCODE: Sort Colors

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于 2014-12-25 21:55:09 发布

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文章标签： string 算法

本文链接：https://blog.youkuaiyun.com/anyicheng2015/article/details/42154285

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本文介绍了一种在不使用库排序函数的情况下，通过两遍扫描算法解决数组中不同颜色元素排序的问题。该算法利用计数排序原理，通过三个指针分别记录0、1、2元素的位置，实现一次遍历完成排序。

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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?

关键点：记住最后一个0,1,2的位置。

class Solution {
public:
    void sortColors(int A[], int n) {
        int endpos0 = -1;
        int endpos1 = -1;
        int endpos2 = -1;
        for(int ii = 0; ii < n; ii ++) {
            if(A[ii] == 0) {
                A[++endpos2] = 2;
                A[++endpos1] = 1;
                A[++endpos0] = 0;
            }
            else if(A[ii] == 1) {
                A[++endpos2] = 2;
                A[++endpos1] = 1;
            }
            else {
                A[++endpos2] = 2;
            }
        }
    }
};