75. Sort Colors

Description

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?

Algorithm

需要一个只扫描一遍的算法。
我的做法是维护当前头部的0和尾部的2，中间都填充1。但是实现的时候就有很多种情形，实现起来有些麻烦。
最后是把各种情况做了转化，来化简情况的。

1. 先得到头部0的序列的末尾
2. 再得到尾部2的序列的头部
3. 在1，2的区间之间扫描，对于当前的i处的color
   是0：如果当前i就是0序列的末尾，可以更新0序列的末尾。否则把0和0末尾后的那个数（应该是1）交换
   是1：满足当前约束，忽略continue
   是2：和末尾的2序列前的一个数交换，i–保持在这里再次考虑，同时更新末尾2序列的头部。（这里把尾部的2序列扩充了）

class Solution {
public:
    // 0 ... 1 ... 2
    void sortColors(vector<int>& nums) {
        int n = nums.size();
        int zero_end = 0;
        while (zero_end < n && nums[zero_end] == 0) zero_end++;

        int two_head = n - 1;
        while (two_head >= zero_end && nums[two_head] == 2) two_head--;

        for (int i = zero_end; i <= two_head; i++) {
            switch (nums[i]) {
                case 0:
                    if (i == zero_end) continue;
                    nums[i] = nums[zero_end];
                    i--;
                    nums[zero_end++] = 0;
                    break;
                case 1:
                    nums[i] = 1;
                    break;
                case 2:
                    nums[i] = nums[two_head];
                    i--;
                    nums[two_head--] = 2;
                    break;
                default:
                    break;
            }
        }
    }
};