Leet Code OJ 136. Single Number [Difficulty: Medium]

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?

分析：
题意是给定一个数组，该数组除了一个元素只出现一次外，其他元素均出现2次，找出这个出现一次的元素。Note中要求具有线性复杂度，也就是O(n)。

本题的考察点是异或运算，解题的关键是利用异或运算的交换律：a ⊕b ⊕ c = a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c，以及a ⊕ a = 0。对数组的全部元素进行异或运算，将里面出现2次的元素交换到一起，这样由于a ⊕ a = 0，最终结果就是那个只出现一次的元素。

代码实现1（时间复杂度O(n)）：

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

代码实现2（时间复杂度O(n^2)）：

public class Solution {public int singleNumber(int[] nums) {int target;for(int i=0;i<nums.length;i++){target=nums[i];boolean found=false;for(int j=0;j<nums.length;j++){if(target==nums[j]&&i!=j){found=true;break;}}if(found==false){return target;}}return 0;}
}

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