Leet Code OJ 136. Single Number

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?

代码实现1：

[code]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：

[code]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;
    }
}