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Given a non-empty array of integers nums
, every element appears twice except for one. Find that single one.
You must implement a solution with a linear runtime complexity and use only constant extra space.
Example 1:
Input: nums = [2,2,1]
Output: 1
Example 2:
Input: nums = [4,1,2,1,2]
Output: 4
Example 3:
Input: nums = [1]
Output: 1
Constraints:
1 <= nums.length <= 3 * 104
-3 * 104 <= nums[i] <= 3 * 104
class Solution {
public:
int singleNumber(vector<int>& nums) {
unordered_map<int,int> a;
for(auto x: nums)
a[x]++;
for(auto z:a)
if(z.second==1)
return z.first;
return -1;
}
};
class Solution:
def singleNumber(self, nums: List[int]) -> int:
xor = 0
for num in nums:
xor ^= num
return xor
public int singleNumber(int[] nums) {
int ans =0;
int len = nums.length;
for(int i=0;i!=len;i++)
ans ^= nums[i];
return ans;
}
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