[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;
    }
};