136. Single Number
- Total Accepted: 173470
- Total Submissions: 331880
- Difficulty: Easy
- Contributors: Admin
Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
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题意：给你一个动态数组，里面的数只有一个数出现过一次其余的数都出现过两次，让你在不另辟空间的情况下用O（n）的复杂度实现，真是神奇的算法，能想到的也只有快排那些O（n*lgn）的东西
实现：因为A和A的异或等于0，而A和0的异或等于A，所以将整个数组遍历每个数异或一遍的结果就是要找的那个数，是不是很神奇
class Solution {
public:
int singleNumber(vector<int>& nums) {
int i = 1;
while(i<nums.size())
{
nums[i]=nums[i-1]^nums[i];
i++;
}
return nums[i-1];
}
};
public:
int singleNumber(vector<int>& nums) {
int i = 1;
while(i<nums.size())
{
nums[i]=nums[i-1]^nums[i];
i++;
}
return nums[i-1];
}
};