268. Missing Number（重要！）

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?

方法一：

求和相减得到的就是消失的。

class Solution {
public:
int missingNumber(vector<int>& nums) {
int len = nums.size();
int sum = (1 + len)*len / 2;
int sum2 = 0;
for (int n : nums){
sum2 += n;
}
return sum - sum2;
}
};

方法二：异或
跟 

137. Single Number II  

260. Single Number III 类似

class Solution {
public:
int missingNumber(vector<int>& nums) {
int res=0;
for(int i=0;i<nums.size();i++){
res^=(i+1)^nums[i];
}
return res;
}
};