[LeetCode] #1 Two Sum

今天开始，每天一道LeetCode！

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

本体采用动态规划的想法，寻找target的数，其实就是index1=target-index2，用map<int,int>存放index1的数和位置，通过hmap.count(target-numbers[i])寻找index1是否存在。时间复杂度O(n)。时间：33ms

代码如下：

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        int i, sum;
        vector<int> results;
        map<int, int> hmap;
        for(i=0; i<numbers.size(); i++){
            if(!hmap.count(numbers[i])){
                hmap.insert(pair<int, int>(numbers[i], i));
            }
            if(hmap.count(target-numbers[i])){
                int n=hmap[target-numbers[i]];
                if(n<i){
                    results.push_back(n+1);
                    results.push_back(i+1);
                    return results;
                }
            }
        }
        return results;
    }
};