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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?
public class Solution { public void sortColors(int[] nums) { int count0=0; int count1=0; int count2=0; int l=nums.length; for(int i=0;i<l;i++){ if(nums[i]==0) count0++; else if(nums[i]==1){ count1++; }else if(nums[i]==2){ count2++; } } for(int j=0;j<l;j++){ if(j<count0){ nums[j]=0; }else if(j<count0+count1){ nums[j]=1; }else{ nums[j]=2; } } } }
一趟遍历算法：转自http://blog.csdn.net/yangliuy/article/details/40584683
public class Solution { public void sortColors(int[] A) { //Implement counting sort int count1 = 0; int count2 = 0; int count0 = 0; for(int i = 0; i < A.length; i++){ if(A[i] == 2){ count2++; } else if(A[i] == 1){ count1++; swap(A, i , i-count2); } else { count0++; swap(A, i, i-count2); swap(A, i-count2, i-count2-count1); } } } public void swap(int[] A, int i, int j){ int temp; temp = A[i]; A[i] = A[j]; A[j] = temp; } }
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原文地址：http://www.cnblogs.com/gonewithgt/p/4572613.html
