标签：c style class blog code java
Single Number
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?
解法一：用map记录每个元素的次数，返回次数为1的元素
class Solution { public: map<int,int> m; int singleNumber(int A[], int n) { for(int i = 0; i < n; i ++) { map<int,int>::iterator it = m.find(A[i]); if(it == m.end()) m.insert(map<int,int>::value_type(A[i], 1)); else it->second = 2; } for(map<int,int>::iterator it = m.begin(); it != m.end(); it ++) { if(it->second == 1) return it->first; } } };
解法二：利用异或操作的结合律。同一数字异或自己结果为0.x^x=0
class Solution { public: int singleNumber(int A[], int n) { int sum = 0; for(int i = 0; i < n; i ++) sum ^= A[i]; return sum; } };
【LeetCode】Single Number (2 solutions),布布扣,bubuko.com
【LeetCode】Single Number (2 solutions)
标签：c style class blog code java
原文地址：http://www.cnblogs.com/ganganloveu/p/3774619.html
