# [LeetCode] 260. Single Number III 单独的数字之三

Given an integer array nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.

You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.

Example 1:

Input: nums = [1,2,1,3,2,5]
Output: [3,5]
Explanation:  [5, 3] is also a valid answer.


Example 2:

Input: nums = [-1,0]
Output: [-1,0]


Example 3:

Input: nums = [0,1]
Output: [1,0]


Constraints:

• 2 <= nums.length <= 3 * 104
• -231 <= nums[i] <= 231 - 1
• Each integer in nums will appear twice, only two integers will appear once.

class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
unsigned int diff = accumulate(nums.begin(), nums.end(), 0, bit_xor<int>());
diff &= -diff;
vector<int> res(2, 0);
for (auto &a : nums) {
if (a & diff) res[0] ^= a;
else res[1] ^= a;
}
return res;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/260

Single Number II

Single Number

https://leetcode.com/problems/single-number-iii/

https://leetcode.com/problems/single-number-iii/discuss/68900/Accepted-C%2B%2BJava-O(n)-time-O(1)-space-Easy-Solution-with-Detail-Explanations

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posted @ 2015-08-19 00:42  Grandyang  阅读(20160)  评论(12编辑  收藏  举报