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]

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?

``````package array;

import static format.PrintArray.printArray;

public class leetcode75SortColors {
/*=====================================================================================*/
/*自想：荷兰国旗问题
* partition过程看成向小于区域和大于区域发货
* 时间复杂度O(n),空间复杂度O(1)
* */
public static void sortColors(int[] nums) {
if (nums == null || nums.length <= 1) return;
partition(nums, 0, nums.length - 1);
}

public static void partition(int[] nums, int lo, int hi) {
int less = -1;
int more = nums.length;
int cur = 0;
while (cur < more) {
if (nums[cur] == 0) {
swap(nums, cur++, ++less);
} else if (nums[cur] == 1) {
cur++;
} else if (nums[cur] == 2) {
swap(nums, cur, --more);
}
}
}

public static void swap(int[] nums, int lo, int hi) {
if (lo == hi) return;
nums[lo] = nums[lo] ^ nums[hi];
nums[hi] = nums[lo] ^ nums[hi];
nums[lo] = nums[lo] ^ nums[hi];
}

/*=====================================================================================*/
public static void main(String[] args) {
int[] nums = new int[]{2,0,2,1,1,0};
printArray(nums);
sortColors(nums);
printArray(nums);
}
}
``````