【HackerRank】Ice Cream Parlor

Sunny and Johnny together have M dollars which they intend to use at the ice cream parlour. Among N flavors available, they have to choose two distinct flavors whose cost equals M. Given a list of cost of N flavors, output the indices of two items whose sum equals M. The cost of a flavor (ci) will be no more than 10000.

Input Format

The first line of the input contains T, T test cases follow. 
Each test case follows the format: The first line contains M. The second line contains the number N. The third line contains N single space separated integers denoting the price of each flavor ci.

Output Format

Output two integers, each of which is a valid index of the flavor. The lower index must be printed first. Indices are indexed from 1 to N.

Constraints

1 ≤ T ≤ 50 
2 ≤ M ≤ 10000 
2 ≤ N ≤ 10000 
1 ≤ ci ≤ 10000 
The prices of two items may be same and each test case has a unique solution.

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又是一个求数组中两个数和正好是m的问题，类似leetcode中的Two Sum问题。

只是在这个问题中，数组中的数是有可能重复的，所以map中的value要用一个arrayList保存。

另外注意找的时候要保持i比在map中找到的坐标小（如代码26行的判断所示），以免重复。

代码如下：

import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while(t-- >0){
            int m = in.nextInt();
            int n = in.nextInt();
            int[] prices = new int[n];
            HashMap<Integer, ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();
            
            for(int i = 0;i < n;i++){
                prices[i]=in.nextInt();
                if(!map.containsKey(prices[i]))
                    map.put(prices[i], new ArrayList<Integer>());
                map.get(prices[i]).add(i);
            }
            
            for(int i = 0;i < n;i++){
                int has = prices[i];
                int toFind = m-prices[i];
                
                if(map.containsKey(toFind)){
                    for(int temp:map.get(toFind)){
                        if(i < temp){
                            System.out.printf("%d %d",i+1,temp+1);
                            System.out.println();
                        }
                    }
                }
            }
        }       
    }
}