【LeetCode】Single Number (2 solutions)

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