LeetCode Sort Colors

最新推荐文章于 2019-05-21 15:41:52 发布

陈琼

最新推荐文章于 2019-05-21 15:41:52 发布

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分类专栏： LeetCode 文章标签： leetcode arraylist 遍历

本文链接：https://blog.youkuaiyun.com/xuyilanqi/article/details/40835615

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本文介绍了一种解决数组中不同颜色对象排序问题的方法，通过一次遍历计算每种颜色的数量，然后根据数量依次填充数组，实现相同颜色的对象相邻排列，且颜色顺序为红、白、蓝。

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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.

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?

方法1：遍历一遍，计算0、1、2的个数，再遍历一遍根据个数赋值

方法2：遍历一遍，如果是0，往前置换；如果是2，往后置换

public class Solution {
    public void sortColors(int[] A) {
        int len=A.length;
        int start=0;
        if(len==0)  //数组为空的情况
        	return;
        int end=len-1;
        for(int i=0;i<len;i++)
        {
        	while(start<len && A[start]==0)  //防止数组越界
        		start++;
        	if(i<start)
        		i=start;
        	while(end>=0 && A[end]==2)  //防止数组越界
    			end--;
        	if(i>end)
        		break;
        	
        	if(A[i]==2)
        	{
        		
        		A[i]=A[end];
        		A[end]=2;
        		end--;
        	}
        	if(A[i]==0)   //不能用else if，需要考虑置换的数值是否为0
        	{
        		
        		A[i]=A[start];
        		A[start]=0;
        		start++;
        	}
        	
        }
    }
}