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.
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：
1 import java.util.*;
2
3 public class Solution {
4 public static void main(String[] args) {
5 Scanner in = new Scanner(System.in);
6 int t = in.nextInt();
7 while(t-- >0){
8 int m = in.nextInt();
9 int n = in.nextInt();
10 int[] prices = new int[n];
11 HashMap<Integer, ArrayList<Integer>> map = new HashMap<Integer, ArrayList<Integer>>();
12
13 for(int i = 0;i < n;i++){
14 prices[i]=in.nextInt();
15 if(!map.containsKey(prices[i]))
16 map.put(prices[i], new ArrayList<Integer>());
17 map.get(prices[i]).add(i);
18 }
19
20 for(int i = 0;i < n;i++){
21 int has = prices[i];
22 int toFind = m-prices[i];
23
24 if(map.containsKey(toFind)){
25 for(int temp:map.get(toFind)){
26 if(i < temp){
27 System.out.printf("%d %d",i+1,temp+1);
28 System.out.println();
29 }
30 }
31 }
32 }
33 }
34 }
35 }
【HackerRank】Ice Cream Parlor