LeetCode（75） Sort Colors

题目

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?

分析

这是一个简单排序问题，所给数组中只有0 ， 1 ， 2 大小的元素，将其由小及大排序即可。

题目明确要求不可用标准库中的排序算法，尽可能使得算法高效。

我们常用的排序算法最优情况下也有 O(nlogn) 的时间复杂度，对待这样一个特殊的数组排序，我们应该采取其他更加高效的方法。

follow up 给出一种方法，分别计算元素0 ， 1 ， 2 的个数，然后对原数组重新赋值，简单高效，下面用该

方法给出程序实现。

AC代码

class Solution {
public:
    void sortColors(vector<int>& nums) {
        if (nums.empty())
            return;

        //初始化一个count数组，count[0] , count[1] , count[2] 分别记录nums中0 ， 1 ， 2出现个数
        vector<int> count(3, 0);

        vector<int>::iterator iter = nums.begin();

        for (; iter != nums.end(); iter++)
        {
            if (*iter == 0)
                count[0]++;
            else if (*iter == 1)
                count[1]++;
            else if (*iter == 2)
                count[2]++;
        }//for

        //对原数组排序
        int i = 0;
        while (i < count[0])
            nums[i++] = 0;
        while (i < (count[0] + count[1]))
            nums[i++] = 1;
        while (i < (count[0] + count[1] + count[2]))
            nums[i++] = 2;

        return;

    }
};

GitHub测试程序源码