p136

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LeetCode P136 Single Number 题解

1.题目：

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

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题意：

输入一个数组，返回这个数组中单出来的那个数，希望只做一次循环不占用其它空间

2.解题思路:

java 位运算的^亦或:二进制每位相同则结果为0，不同则结果为1。
由定义可以知道:0^n=n,n^n=0;

3.代码

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package leetcode;

import java.util.Scanner;


/**
 * 
 * @author ZhangMengRou
 *
 */
public class p292 {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String a = in.nextLine();
		String[] list = a.split(" ");
		int[] al = new int[list.length];
		for (int i=0;i<list.length;i++)
		{
			al[i]=Integer.parseInt(list[i]);
		}
		int ans = singleNumber(al);
		System.out.print(ans);
	}

	public static int singleNumber(int[] nums) {
		
		int as = 0;
		for (int i = 0; i < nums.length; i++) {
			as = as ^ nums[i];
		}
		return as;
        
    }

}


4.一些总结：

有时候位运算是一个不错的选择~不过好像就知道这一种符合要求的方法，
希望可以找到新的方法补充。