LeetCode:1. Two Sum

问题描述:Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.
给定一个数组和一个目标值,找到两个数组相加的和为目标值.只有一个解,不能够使用两次相同的元素.

LZ的想法比较简单,两次循环,好暴力…

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int> indice_a;
        for(int i = 0; i<nums.size()-1; i++){
            for(int j = i+1; j<nums.size(); j++){
                if(nums[i]+nums[j]==target)
                {
                    indice_a.push_back(i);
                    indice_a.push_back(j);
                    break;
                }
            }
        }
     return indice_a;
    }

};

结果测试时间是216ms!

为了减少runtime,也即时间复杂度

使用哈希表,只要10ms!

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
       std::unordered_map<int, int> mp;

        std::vector<int> ans;

        //  用哈希表来做
        for(int i = 0; i < nums.size(); i ++)    //  对于数组中的每个数据
        {
            //  判断target - numbers[i]在不在
            if(mp.count(target - nums[i]))   //  如果在里面   
            {
                //  那么numbers中就存在两个数和为target
                ans.push_back(mp[target - nums[i]] ); 
                ans.push_back(i );
                break;
            }

            if(mp.count(nums[i]) != 1)
            {
                mp[nums[i]] = i;
            }
        }
        return ans;
    }

};

代码的魅力在于此^_^