# [leetcode] Sort Colors

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.

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?

oneStart表示排序后1开始的位置，twoStart表示排序后2开始的位置。0从0处开始，2到i处结束。

```class Solution {
public:
void sortColors(int A[], int n) {
int i = 0;
int oneStart = 0;
int twoStart = 0;
while(i < n){
if(A[i] == 0){
A[oneStart++] = 0;
if(oneStart > twoStart){
twoStart++;
}else{
A[twoStart++] = 1;
}
if(twoStart > i){
;
}
else{
A[i] = 2;
}
}
else if(A[i] == 1){
A[twoStart++] = 1;
if(twoStart > i){
;
}
else{
A[i] = 2;
}
}
i++;
}
}
};```

An quick sort approach

```class Solution {
public:
void sortColors(vector<int>& nums) {
_qsort(nums, 0, nums.size() - 1);
}
//Both nums[start] and nums[end] are availiable
void _qsort(vector<int>& nums, int start, int end){
if(start >= end) return;
int pivot = median(nums, start, end);
int i = start;
int j = end - 1;
while(i < j){
while(nums[++i] < pivot)
;
while(nums[--j] > pivot)
;
if(i < j){
swap(nums[i], nums[j]);
}
}
swap(nums[end - 1], nums[i]);
_qsort(nums, start, i - 1);
_qsort(nums, i + 1, end);
}
int median(vector<int>& nums, int start, int end){
int mid = (start + end) / 2;
if(nums[start] > nums[mid]){
swap(nums[start], nums[mid]);
}
if(nums[mid] > nums[end]){
swap(nums[mid], nums[end]);
}
if(nums[start] > nums[mid]){
swap(nums[start], nums[mid]);
}
swap(nums[mid], nums[end - 1]);
return nums[end - 1];
}
};```

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