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  for adding this problem and creating all test cases.

 

这题需要技巧，利用异或。

先将nums的数作异或得到x1；

将0-n+1都作异或得到x2；

missing number 就是x1^x2。

为什么呢？

假设缺少的值为A，则x2=x1^A，那么x1^x2=x1^(x1^A)=(x1^x1)^A=0^A=A，得证。

1 class Solution { 2 public: 3     int missingNumber(vector
     
      & nums) { 4         int result; 5         int x1=nums[0]; 6         for(int i=1;i