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?
首先对0~ n内数字做异或,在对数组内数字做异或。等价与在Single Number 1,在一堆重复数字中找1个。
class Solution {
public:
int missingNumber(vector<int>& nums) {
int ret = 0;
for(int i=0; i<=nums.size(); ++i){
ret ^= i;
}
for(int i=0; i<nums.size(); ++i){
ret ^= nums[i];
}
return ret;
}
};