# 75. Sort Colors

## problem description

Given an array with _n_ objects colored red, white or blue, sort them [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) 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?

## algorithm thought

典型的计数排序，计数排序时间复杂度是 O(n)，但是适用于数字比较少的情况。这里一共就3个数字，是比较适合的。

最简单的一种方法是，两次遍历出结果，第一次遍历得到所有数的count，第二次遍历对数组赋值。

能否一次遍历出结果呢？是可以的，使用两个标志位标志0，1的结尾。遍历数组，如果为0，swap交换到0的末尾，如果为1，swap交换到1的末尾

## code

```cpp
class Solution {
public:
    void sortColors(vector<int>& nums) {
        int pos_1=0,pos_2=0;
        for(int i=0;i<nums.size();++i){
            if(nums[i]==1){
                swap(nums[i],nums[pos_2++]);
            }
            else if(nums[i]==0){
                swap(nums[i],nums[pos_1++]);
                if(pos_1-1!=pos_2)
                    swap(nums[i],nums[pos_2++]);
                else
                    pos_2++;
            }
        }
    }
};
```

## algorithm analysis

一次遍历数组，时间复杂度O(n)
