leetcode136 Single Number
题意:数组中每个数字都出现了两次,只有一个出现一次,找出这个数
思路:很明显不能从头到位遍历来找,首先是超时的原因,再次就是这样很没意思·····但是却没想到什么好办法,因为不了解按位异或(XOR)。异或就是相同的两个数结果为0,不同的为1。根据交换律我们知道,数组中两两异或的结果就剩最后那一个落单的数了,因为两个相同的数异或结果为0。所以其实这题就很简单了,只是如果没想到或者根本不知道异或就无法好好解题了。
代码:
int singleNumber(vector<int>& nums) {
int ans = ;
for(int n : nums)
{
ans ^= n;
}
return ans;
}
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