# 75. Sort Colors

75. Sort Colors

Given an array with n objects colored red, white or blue, sort them in-place 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.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

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 a one-pass algorithm using only constant space?

  颜色有0，1，2三种，只需要把数组中为0的元素置于数组的前部，数组中为2的元素置于数组的后部，放置好0与2之后，1会被放置在数组的中部

class Solution {
public:
    void sortColors(vector<int>& A) {
        int size = A.size();
        int start = 0;
        int end = size - 1;
        for(unsigned i = 0; i<=end; i++)//i<=end:在重复查找元素之前停止
        {
            int temp;
            while(A[i] == 2 && i<end)//i<end：避免进入死循环[2,0,2,1,1,0]
            {//交换
                temp = A[end];
                A[end] = A[i];
                A[i] = temp;
                end--;
            }
            while(A[i] == 0  && i>start)//i>start：避免进入死循环[2,0,2,1,1,0]
            {
                temp = A[start];
                A[start] = A[i];
                A[i] = temp;
                start++;
            }

        }
    }
};

注意程序中的两个while循环位置不能对换，因为索引i按0->size-1的方向来遍历数组的，在它已经遍历的数组元素中肯定不会有2存在（有的话也已经被移到最后了），但是在它没有遍历的数组元素中，可能会有0的存在比如[1,2,0]，此时如果while(A[i] == 0 && i>start)在前，而while(A[i] == 2 && i<end)在后，输出为[1,0,2]