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## Problem

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]


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?

## Code

class Solution {
public void sortColors(int[] nums) {
int point0 = 0;
int point2 = nums.length - 1;
int i = 0;

while(i <= point2){
if(nums[i] == 0){
// 换回来的一定不是2, 因为之前i已经遍历过了
swap(nums, point0, i);
i++;
point0++;
} else if(nums[i] == 1){
i++;
} else {
// 有可能换回来的是0
// 所以不能i++
swap(nums, point2, i);
point2--;
}
}
}

public void swap(int[] nums, int a, int b){
int tmp = nums[a];
nums[a] = nums[b];
nums[b] = tmp;
}
}