## Sort Colors

### 描述

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.

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?

### 代码1

// Sort Colors
// Counting Sort
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public void sortColors(int[] nums) {
int[] counts = new int[3]; // 记录每个颜色出现的次数

for (int i = 0; i < nums.length; i++)
counts[nums[i]]++;

for (int i = 0, index = 0; i < 3; i++)
for (int j = 0; j < counts[i]; j++)
nums[index++] = i;

}
}

### 代码2

// Sort Colors
// 双指针，时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public void sortColors(int[] A) {
// 一个是red的index，一个是blue的index，两边往中间走
int red = 0, blue = A.length - 1;

for (int i = 0; i < blue + 1;) {
if (A[i] == 0)
swap (A, i++, red++);
else if (A[i] == 2)
swap(A, i, blue--);
else
i++;
}
}
private static void swap(int[] A, int i, int j) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
}

### 代码3

// Sort Colors
// 重新实现 partition()
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public void sortColors(int[] nums) {
partition(nums, partition(nums, 0, nums.length, new EqualTo(0)),
nums.length, new EqualTo(1));
}
private static int partition(int[] nums, int begin, int end, EqualTo predicate) {
int pos = begin;

for (; begin != end; ++begin)
if (predicate.apply(nums[begin]))
swap(nums, begin,pos++);

return pos;
}
static class EqualTo {
private final int target;
public EqualTo(int target) {
this.target = target;
}
public boolean apply(int x) {
return x == target;
}
}
private static void swap(int[] nums, int i, int j) {
int tmp = nums[i];
nums[i] = nums[j];
nums[j] = tmp;
}
}