Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
class Solution {
public:
int singleNumber(vector<int>& nums) {
for(int i=1;i<nums.size();i++){
nums[0]^=nums[i];
}
return nums[0];
}
};
要求O（n）並且不用額外的變量。 非常巧妙的技巧。相同數取異或爲0，所0和所有數的異或爲本身，所以最後剩下的就是Single Number。