136. Single Number



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?

解法一：对数组排序，判断如果和下一个相同，向前移动两位，直到找到和下一个不同的，return这个数；最后一个需特殊处理

public class Solution {    public int singleNumber(int[] nums) {        if(nums==null||nums.length==0)            return -1;        Arrays.sort(nums);        int i=0;        while(i<nums.length-1)        {            if(nums[i]==nums[i+1])            {                i=i+2;            }            else                return nums[i];        }        return nums[nums.length-1];    }}

解法二：0^N=N;N^N=0;只需要遍历数组，逐个异或，出现两次的都是0，最后遍历完的结果就是只出现一次的那个元素N

public int singleNumber(int[] nums) {    int ans =0;       int len = nums.length;    for(int i=0;i!=len;i++)        ans ^= nums[i];    return ans;}