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[leetcode]Sort Colors

时间:2014-08-12 00:09:03      阅读:154      评论:0      收藏:0      [点我收藏+]

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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.

click to show follow up.

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?

算法思路:

思路1: 两边扫描,题中已经给了提示,开个hash[3],记录个数,然后再次填充array

代码如下:

 1 public class Solution {
 2     public void sortColors(int[] A) {
 3         int[] hash = new int[3];
 4         for(int tem : A){
 5             hash[tem]++;
 6         }
 7         int index = 0;
 8         for(int i = 0; i < 3;i++){
 9             while(hash[i] != 0){
10                 A[index++] = i;
11                 hash[i]--;
12             }
13         }
14     }
15 }

思路2: 

双指针法。(其实是algorithms里讲的的quicksort中的three-way-partition部分,详见这里

  •   两指针一前一后分别指向1的开始元素和1的结束元素。

代码如下:

public class Solution {
    public void sortColors(int[] a) {
        if (a == null || a.length < 2) return;
        int end = a.length - 1,begin = 0;
        int i = 0;
        while(i <= end){
            if(a[i] == 1){
                i++;
            }else if(a[i] == 2){
                swap(a, end--, i);
            }else{
                swap(a, begin++, i++);
            }
        }
    }
    
    private void swap(int[] a,int i ,int j){
        int tem = a[i];
        a[i] = a[j];
        a[j] = tem;
    }
    
}

 

[leetcode]Sort Colors,布布扣,bubuko.com

[leetcode]Sort Colors

标签:style   blog   http   color   os   io   strong   for   

原文地址:http://www.cnblogs.com/huntfor/p/3905739.html

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