Given an array withnobjects 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?
思路：如果空间上不做要求，这题还是比较简单的。可以再声明一个m,n矩阵，然后搜索原矩阵，是0，置于首位，是2置于末尾，中间置为1.
代码如下：
public class Solution {public void sortColors(int[] nums) {int[] a = new int[nums.length];a = Arrays.copyOf(nums, nums.length);int i = 0;int j = nums.length – 1;for(int k = 0; k < nums.length; k++){if(a[k] == 0){nums[i++] = a[k];}else if(a[k] == 2){nums[j–] = a[k];}}while(i <= j){nums[i++] = 1;}}}但是如果常数空间的话，，就有点难度，网上通用的解法如下（可以对n个数排序）：
public class Solution {public void sortColors(int[] nums) {int i = -1,j = -1,k = -1;for(int m = 0; m < nums.length; m++){if(nums[m] == 0){nums[++k] = 2;nums[++j] = 1;nums[++i] = 0;}else if(nums[m] == 1){nums[++k] = 2;nums[++j] = 1;}else{nums[++k] = 2;}}}}
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