第十一周算法分析与设计： Sort Colors

问题描述：

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.
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?

题目来自此处

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解法思路：
不给用库函数……而且函数无返回值即是要在原数组内进行处理。
可以按题目提示的Follow up的思路进行处理——统计原数组中0，1，2的个数，然后清空数组，分别把这三个数依次写进数组内。不用返回~

class Solution {
public:
    void sortColors(vector<int>& nums) {
        int size = nums.size();
        int num[3] = {0};
        for(int i=0;i<size;i++) {
            if(nums[i] == 0)
                num[0]++;
            else if(nums[i] == 1)
                num[1]++;
            else
                num[2]++;
        }
        nums.clear();
        for(int j=0;j<3;j++) {
            for(int k=0;k<num[j];k++) {
                nums.push_back(j);
            }
        }
    }
};

算法的时间和空间复杂度都为$O(N)$。但是上面的解法却不能在一次扫描内解决，想在一次扫描内解决的方法可以采用内部元素交换的方式进行数组的处理咯。

又是一道比较水的题目嘿……