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  • [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。

    1 class Solution {
    2 public:
    3     int singleNumber(int A[], int n) {
    4         int temp;
    5         for(int i=0;i!=n;i++)
    6             temp=temp^A[i];
    7         return temp;
    8     }
    9 };

     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。

     1 class Solution {
     2 public:
     3     int singleNumber(int A[], int n) {
     4         int one=0,two=0,three=0;
     5         for(int i=0;i!=n;i++){
     6             three = A[i] & two;
     7             two=two | (one & A[i]);
     8             one = one | A[i];
     9             one = one & ~three;
    10             two = two & ~three;
    11         }
    12         return one;
    13     }
    14 };
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  • 原文地址:https://www.cnblogs.com/desp/p/4333878.html
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