[leetcode-268-Missing Number]

Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

For example,
Given nums = [0, 1, 3] return 2.

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

思路：

The basic idea is to use XOR operation. We all know that a^b^b =a, which means two xor operations with the same number will eliminate the number and reveal the original number.
In this solution, I apply XOR operation to both the index and value of the array. In a complete array with no missing numbers, the index and value should be perfectly corresponding( nums[index] = index), so in a missing array, what left finally is the missing number.

int missingNumber(vector<int>& nums)
     {
         int ret = nums.size();
         for (int i = 0; i < nums.size();i++)
         {
             ret = ret ^ i ^ nums[i];
         }
         return ret;
     }

参考：

https://discuss.leetcode.com/topic/24535/4-line-simple-java-bit-manipulate-solution-with-explaination