136. Single Number

136. Single Number

题目

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?

思路

遍历数组的过程中判断该数字是否在字典中存在，不存在，插入，存在删除，最后剩在字典中的最是答案。时间复杂度为O（n）

代码

    int singleNumber(vector<int>& nums) {      unordered_map<int ,int > map1{};        for(auto iter=nums.begin();iter!=nums.end();iter++)        {            if(map1.count((*iter)))                map1.erase((*iter));            else                map1[*iter]=(iter-nums.begin());        }        for (auto i = map1.begin(); i!=map1.end() ; ++i) {            cout<<(i->first)<<ends;        }        if(map1.size())            return (map1.begin())->first;        else            return 0;    }