【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();
                         }
                     }
                 }
             }
         }
     }
 }