leetcode 练习1 two sum

　　　　leetcode 练习1  two sum

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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.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].
解法1：暴力破解法: O(n^2) runtime, O(1) space – Brute force: The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through

the rest of array, its runtime complexity is O(n^2).

 

解法2:使用HashMap。把每个数都存入map中，然后再逐个遍历，查找是否有 target – nums[i]。  O(n) runtime  O(n) space,

vector<int> twoSum(vector<int>& nums, int target){

       vector<int> vec;

       map<int,int> m;

       for (int i = 0; i < nums.size(); i++)

       {

              if(m.find(target - nums[i]) != m.end())

              {

                     vec.push_back(m[target - nums[i]] );

                     vec.push_back(i);

                     break;

              }

              m[nums[i]] = i;

       }

       return vec;

}

调用：

       int a[6]={2,7,1,8,9};

       vector<int> vec(a,a+5);

       vector <int> vect= twoSum(vec,15);

       return 1;