leetcode【1】2021-07-28

1.两数之和

思路1：暴力遍历[o(n^2)]

#include <vector>
#include <map>
using namespace std;
class Solution {
    
    
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    
    
        vector<int> res;
        for (int i = 0; i < nums.size(); i++)
        {
    
    
            for (size_t j = i+1; j < nums.size(); j++)
            {
    
    
                /* code */
                if(nums[j]==target-nums[i])//keypoint
                {
    
    
                    res.push_back(i);
                    res.push_back(j);
                }
            }
            
        }
        return res;
    }
};

思路2：使用哈希表[o(n)]
关键点在于使用map做hash存储，相比暴力的2次循环，在第二次循环被替换为o（1）。

#include <vector>
#include <unordered_map>
using namespace std;
class Solution {
    
    
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    
    
        vector<int> res;
        unordered_map<int,int> mapT;
        for (int i = 0; i < nums.size(); i++)
        {
    
    
            unordered_map<int,int>::iterator iter=mapT.find(target-nums[i]);
            if(iter!=mapT.end())
            {
    
    
                res.push_back(iter->second);
                res.push_back(i);
            }
            else
            {
    
    
                mapT.insert(pair<int,int>(nums[i],i));//keypoint
            }
        }
        return res;
    }
};