代码拉取完成,页面将自动刷新
同步操作将从 doocs/leetcode 强制同步,此操作会覆盖自 Fork 仓库以来所做的任何修改,且无法恢复!!!
确定后同步将在后台操作,完成时将刷新页面,请耐心等待。
Given a non-empty array of integers nums
, every element appears twice except for one. Find that single one.
Follow up: Could you implement a solution with a linear runtime complexity and without using extra memory?
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:
def singleNumber(self, nums: List[int]) -> int:
res = 0
for num in nums:
res ^= num
return res
class Solution {
public int singleNumber(int[] nums) {
int res = 0;
for (int num : nums) {
res ^= num;
}
return res;
}
}
/**
* @param {number[]} nums
* @return {number}
*/
var singleNumber = function (nums) {
let res = 0;
for (let num of nums) {
res ^= num;
}
return res;
};
func singleNumber(nums []int) int {
res := 0
for _, v := range nums {
res ^= v
}
return res
}
class Solution {
public:
int singleNumber(vector<int>& nums) {
int res = 0;
for (auto num : nums) {
res ^= num;
}
return res;
}
};
此处可能存在不合适展示的内容,页面不予展示。您可通过相关编辑功能自查并修改。
如您确认内容无涉及 不当用语 / 纯广告导流 / 暴力 / 低俗色情 / 侵权 / 盗版 / 虚假 / 无价值内容或违法国家有关法律法规的内容,可点击提交进行申诉,我们将尽快为您处理。