[leetcode][hash] Single Number II

题目：

Given an array of integers, every element appears three times except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

class Solution {
public:
//哈希。在哈希表中找到一样的，如果元素个数为2，删除之，否则个数加1；在哈希表中找不到一样的，插入该元素并只个数为1
//时间复杂度O(n),空间复杂度O(n)
int singleNumber(vector<int>& nums) {
if(nums.empty()) return 0;
map<int, int> hashTable;//key表示元素值，val表示元素个数
for(int i = 0; i < nums.size(); ++i){
map<int, int>::iterator iter = hashTable.find(nums[i]);
if(iter != hashTable.end()){//找到了
if(iter->second == 2) hashTable.erase(nums[i]);
else iter->second++;
}
else hashTable.insert(make_pair(nums[i], 1));
}
return hashTable.begin()->first;
}
};

希望以后可以找到效率更高的算法