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.

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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：count sort，count一遍，然后赋值；

public class Solution {    public void sortColors(int[] nums) {        if(nums == null || nums.length == 0) return;        int redcount = 0; int whitecount = 0;        for(int i=0; i<nums.length; i++){            if(nums[i] == 0){                redcount++;            } else if(nums[i] == 1){                whitecount++;            }        }        for(int i=0; i<nums.length; i++){            if(i<redcount){                nums[i] = 0;            } else if(i>=redcount && i<redcount+whitecount){                nums[i] = 1;            } else {                nums[i] = 2;            }        }    }}

思路：三指针，i, j, k. 目标是 0 ，1 ，2

用i来保持0的位置，k来保持2的位置，j来做扫描运动，如果j遇见0，换到前面去，如果遇见2换到后面去。

注意，如果换到前面的情况，i可以++，j也可以++，因为换回来的只可能是1，不可能是别的情况，因为前面的i部分已经被扫描过，2已经到后面去了。

j遇见2，与后面进行互换的时候，j不能++，因为不知道换回来的是0，还是1，还得进行一次判断。这也是这个题目的考点。

public class Solution {     public void sortColors(int[] nums) {         if(nums == null || nums.length == 0) return;         int i = 0; int j = 0; int k = nums.length-1;         while(j <= k){             if(nums[j]==1){                 j++;             } else if(nums[j] == 0) {                 swap(nums, i, j);                 i++;                 j++;             } else {                 swap(nums, j, k);                 k--;             }         }     }          public void swap(int[] nums, int i, int j){         int temp = nums[i];         nums[i] = nums[j];         nums[j] = temp;     } }