TwoSum

题目：

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

给出一个整型数组，给出一个整数，求出数组中两个数相加等于该整数的下标。

解法：第一个方法是二级遍历，算法复杂度是O（n^2）

代码如下：

vector<int> twoSum(vector<int> &numbers, int target){    vector <int> results;    int j, i;    for (i = 0; i <numbers.size()-1;i++)    for (j = i + 1; j < numbers.size(); j++)            if (numbers[j]+numbers[i]==target){            results.push_back(i+1);            results.push_back(j+1);        }        return results;}

O（n^2）算法复杂度的肯定是不能满足需求的：

因此考虑用哈希表来实现，直接用C++中的map关联容器。

其代码如下：

vector<int> twoSum(vector<int> &numbers, int target){vector <int>results;map<int, int>hmap;int i;for (i = 0; i < numbers.size(); i++){if (!hmap.count(numbers[i]))hmap.insert(pair<int, int>(numbers[i], i+1));//对MAP初始化}for (i = 0; i < numbers.size(); i++){//cout << "a" << endl;if (hmap.count(target - numbers[i]))if( (i + 1)<(hmap[target - numbers[i]])){results.push_back(i + 1);results.push_back(hmap[target - numbers[i]]);}}return results;}

通过这个题，学习到对vector容器和map关联容器的操作，