# 75. Sort Colors (1)

[Link](https://leetcode.com/problems/sort-colors/submissions/)

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?

**Solution**

這題思路是設計兩個pointer分別指向0的下個位置(indexOf0)跟2的上個位置(indexOf2)。然後iteration每個element如果是0 就swap到indexOf0，如果是2就就swap到indexOf2.

![](https://2628390332-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-M0RnkuvOeqe0clAIN1b%2F-M3KUV0BeHfHXNM9wTuc%2F-M3Kd5nR2WS0BXHc0-cK%2Fimage.png?alt=media\&token=543c84dd-fd36-4c85-a6b3-fa5404b6c634)

值得注意的是i++何時要觸發，若是indexOf0==i ，則i++，因為經過這輪i已經處理過了，不用再來一次。

```
class Solution {
    public void sortColors(int[] nums) {
        int len = nums.length;
        int indexOf2 = len - 1;
        int indexOf0 = 0;
        int i = 0;
        while(i <= indexOf2){
            if(nums[i] == 2){
                nums[i] = nums[indexOf2];
                nums[indexOf2] = 2;
                indexOf2--;
            }
            else if(nums[i] == 0){
                nums[i] = nums[indexOf0];
                nums[indexOf0] = 0;
                if(i == indexOf0) i++;
                indexOf0++;
            }else{
                i++;
            }
            
        }
        
    }
}
```
