136. Single Number

Given a non-empty array of integers nums, every element appears twice except for one. Find that single one.

You must implement a solution with a linear runtime complexity and use only constant extra space.

位运算，对所有数值做二进制异或运算。
两个同样的值异或运算会等于0，最后和与单独的数字相等。

class Solution {
    public int singleNumber(int[] nums) {
        int ans = 0;
        for(int num : nums){
            ans = ans ^ num;
        }
        return ans;
    }
}

排序，然后遍历数组，如果第i个值不等于第i+1个则返回。

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