点击打开链接
/*
* @lc app=leetcode.cn id=260 lang=cpp
*
* [260] 只出现一次的数字 III
*/
// @lc code=start
class Solution {
public:
vector<int> singleNumber(vector<int>& nums) {
int diff = 0;
for (int n: nums)
diff ^= n;
int bit = diff&(-diff); // 取二进制最后一个1
int x = 0;
for (int n: nums)
{
if (n&bit) // 这一位为1
x ^= n;
}
return {x, diff^x};
}
};
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