[Leetcode]Single Number && Single Number II

Given an array of integers, every element appears twice except for one. Find that single one.

非常简单的一道题。

直接相异或剩下的那个数就是答案。原理是两个相等的数异或的值为0。

class Solution {
public:
int singleNumber(int A[], int n) {
int temp;
for(int i=0;i!=n;i++)
temp=temp^A[i];
return temp;
}
};

Given an array of integers, every element appears three times except for one. Find that single one.

用好位运算，多看看与或非到底能做什么。

class Solution {
public:
int singleNumber(int A[], int n) {
int one=0,two=0,three=0;
for(int i=0;i!=n;i++){
three = A[i] & two;
two=two | (one & A[i]);
one = one | A[i];
one = one & ~three;
two = two & ~three;
}
return one;
}
};