[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=;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=,two=,three=;
         for(int i=;i!=n;i++){
             three = A[i] & two;
             two=two | (one & A[i]);
             one = one | A[i];
             one = one & ~three;
             two = two & ~three;
         }
         return one;
     }
 };