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?
s思路: 1. 这道题遇到过! 不用内存,那就是一直做xor运算,相同的数xor等于0,最后只剩下the single number. 2. 这道题就是利用两个数异或的性质,基本也就是一道数学题!
class Solution {
public:
int singleNumber(
vector<int>& nums) {
int res=
0;
for(
int i=
0;i<nums.size();i++){
res^=nums[i];
}
return res;
}
};
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