Leecode -- TwoSum

question:

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

思路一：遍历两次， 时间复杂度O(n2)  O(1)的空间；

思路二：利用hash table；直接遍历一边，看是否有key(target-x)存在； 时间复杂度O(n),space O(n)

code:

class Solution {
public:
　　vector<int> twoSum(vector<int> &numbers, int target) {
　　　　map<int,int> sum;
　　　　vector<int> res;
　　　　for(int i=0;i<numbers.size();i++){
　　　　　　int x = numbers[i];
　　　　　　if(sum.count(target-x)){
　　　　　　　　res.push_back(sum[target-x] + 1);
　　　　　　　　res.push_back(i + 1);
　　　　　　　　return res;
　　　　　　}else{
　　　　　　　　sum[x] = i;
　　　　　　}
　　　　}
　　}
};