LeetCode OJ平台Sort Colors讨论主题算法

原题如下面，这意味着无序排列（由0,1,2组成）。一号通。组织成若干阵列0-几个1-几个2这样的序列。

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?

想了一段时间，未果，于是前往Discuss区域围观。

帖子网址点击打开链接

网友xianlei给出了一个精妙的方法。我理解之后，写出代码例如以下，由于有具体的凝视。就不啰嗦了。

在草纸上模拟一下，或者单步调试一下。就会明确这种方法的精妙之处！

！

public class Solution {
	static int[] A = {1,0};
    public static void sortColors(int[] A) {
    	int i = 0;//last index of array 0
    	int j = 0;//last index of array 1
    	int k = 0;//last index of array 2
    	for(int m = 0; m < A.length; m ++){
    		if(A[m] == 0){ // add 2 to the end of 2 array, replace the 1st 2 with 1, then insert 0 into 0 array,
    			A[k++] = 2;
    			A[j++] = 1;
    			A[i++] = 0;
    		}
    		else if(A[m] == 1){// add 2 to the end of 2 array, then insert 1, at the location of the 1st 2
    			A[k++] = 2;
    			A[j++] = 1;
    		}
    		else{//add 2 to the end of 2 array
    			A[k++] = 2;
    		}
    	}
    }
    public static void main(String args[]){
    	sortColors(A);
    }
}

网友lchen77提供了还有一个非常好的思路，我的实现代码例如以下。

由于题目要求依照0,1,2的顺序来排序，所以遇到0。与前面的A[k++]交换。遇到2与A[j--]交换。

须要注意的是，A[k]要么等于0，要么等于1，所以0与A[k++]交换之后。i不用--，整个前面的序列是合法的。

可是2与A[j--]交换之后。i要--。由于从数组尾部随便一个数字与当前的2换完之后，必须还要验证。以防不合法。

并且，循环的终止条件是i,j相遇，相遇的时候，还要运行一次，即循环终止条件是i>j

public class Solution {
	public static void sortColors(int[] A){
		int k = 0; //index for 0 array;
		int j = A.length-1;//index for 2 array
		for(int i = 0; i <= j;i ++){
			if(A[i] == 1){
				continue;
			}
			else if(A[i] == 0){
				A[i] = A[k];
				A[k++] = 0;
			}
			else{
				A[i] = A[j];
				A[j--] = 2;
				i --;
			}
		}
	}
}

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