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?

two-pass solution

1 public class Solution { 2     public void sortColors(int[] A) { 3         // Start typing your Java solution below 4         // DO NOT write main() function 5         int len = A.length; 6         Map
     
       times = new HashMap
      
       (); 7         for(int i = 0; i < len; i++){ 8             if(times.get(A[i]) != null){ 9                 times.put(A[i], times.get(A[i]) + 1);10             } else {11                 times.put(A[i], 1);12             }13         }14         15         //A = new int[len];16         int red = 0, white = 0, blue = 0;17         18         if(times.get(0) != null)19             red = times.get(0);20         21         if(times.get(1) != null)22             white = times.get(1);23         24         if(times.get(2) != null)25             blue = times.get(2);26             27         for(int i = 0; i < len; i++){28             if(i < red){29                 A[i] = 0;30             } else if(i >= red && i < red + white){31                 A[i] = 1;32             } else {33                 A[i] = 2;34             }35         }36     }37 }
      
     

 one pass

维护两个指针：redIdx，blueIdx，从头开始扫描数组直到blueIdx(包括blueIdx)

1.A[i]==0时，将A[i]与A[redIdx]交换，redIdx++，i++

2.A[i]==2时，将A[i]与A[blueIdx]交换，blueIdx--，

3.A[i]==1时，i++

1 public class Solution { 2     public void sortColors(int[] A) { 3         // Start typing your Java solution below 4         // DO NOT write main() function 5         int len = A.length; 6         int i = 0, redIdx = 0, blueIdx = len - 1; 7         while(i < blueIdx + 1){ 8             if(A[i] == 0){ 9                 swap(A, i, redIdx);10                 redIdx ++;11                 i ++;12             } else if(A[i] == 2){13                 swap(A, i, blueIdx);14                 blueIdx --;15             } else{16                 i++;17             }18         }19     }20     21     public void swap(int[] A, int i, int idx){22         int tmp = A[i];23         A[i] = A[idx];24         A[idx] = tmp;25     }26 }