【LeetCode】1. Two Sum

1. Two Sum

Easy

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.

Example:

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

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

1. 使用HashMap记录元素和所在位置。

2. 检索vector中每一个元素值m[i]，通过HashMap查找是否有target-m[i]的元素存在且不与本身重复。

3. 如果存在，记录并退出。

#include<iostream>
#include<vector>
#include<map>
using namespace std;
vector<int> twoSum(vector<int>& nums, int target) {
	map<int,int> m;    //数值+位置 
	vector<int> ans;
	for(int i=0;i<nums.size();i++){
		m[nums[i]]=i;
	}
	for(int i=0;i<nums.size();i++){
		int t = target - nums[i];
		if(m.count(t)&&m[t]!=i){
			ans.push_back(i);
			ans.push_back(m[t]);
			break;
		}
	}
	return ans;
}
int main(void){
	int n,elem,target;vector<int> input;
	while(cin>>n>>target){
		if(!input.empty()) input.pop_back();
		for(int i=0;i<n;i++){
			cin>>elem;
			input.push_back(elem);
		}
		vector<int> ans = twoSum(input,target);
		cout<<ans[0]<<" "<<ans[1]<<endl;
	}
	return 0;
}