标签：ages int Ages age request Appropriate hm 825 friend
原题链接在这里：https://leetcode.com/problems/friends-of-appropriate-ages/
题目：
Some people will make friend requests. The list of their ages is given and ages[i] is the age of the ith person.
Person A will NOT friend request person B (B != A) if any of the following conditions are true:
age[B] <= 0.5 * age[A] + 7age[B] > age[A]age[B] > 100 && age[A] < 100
Otherwise, A will friend request B.
Note that if A requests B, B does not necessarily request A. Also, people will not friend request themselves.
How many total friend requests are made?
Example 1:
Input: [16,16] Output: 2 Explanation: 2 people friend request each other.
Example 2:
Input: [16,17,18] Output: 2 Explanation: Friend requests are made 17 -> 16, 18 -> 17.
Example 3:
Input: [20,30,100,110,120] Output: Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.
Notes:
1 <= ages.length <= 20000.1 <= ages[i] <= 120.
题解：
Accumlate the frequency of different ages.
If age a and age b could send request, and a != b, then res += a freq * b freq.
If a == b, since no one could send friend request to themselves, the request is a freq * (a freq - 1). Send friend request to other people with same age.
Time Complexity: O(n^2). n = ages.length.
Space: O(n).
AC Java:
1 class Solution {
2 public int numFriendRequests(int[] ages) {
3 if(ages == null || ages.length == 0){
4 return 0;
5 }
6
7 HashMap<Integer, Integer> hm = new HashMap<>();
8 for(int age : ages){
9 hm.put(age, hm.getOrDefault(age, 0) + 1);
10 }
11
12 int res = 0;
13 for(int a : hm.keySet()){
14 for(int b : hm.keySet()){
15 if(couldSendRequest(a, b)){
16 res += hm.get(a) * (hm.get(b) - (a == b ? 1 : 0));
17 }
18 }
19 }
20
21 return res;
22 }
23
24 private boolean couldSendRequest(int a, int b){
25 return !(b <= a*0.5 + 7 || b > a || (b > 100 && a < 100));
26 }
27 }
标签：ages,int,Ages,age,request,Appropriate,hm,825,friend 来源： https://www.cnblogs.com/Dylan-Java-NYC/p/12015827.html
本站声明： 1. iCode9 技术分享网（下文简称本站）提供的所有内容，仅供技术学习、探讨和分享； 2. 关于本站的所有留言、评论、转载及引用，纯属内容发起人的个人观点，与本站观点和立场无关； 3. 关于本站的所有言论和文字，纯属内容发起人的个人观点，与本站观点和立场无关； 4. 本站文章均是网友提供，不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属；如您发现该文章侵犯了您的权益，可联系我们第一时间进行删除； 5. 本站为非盈利性的个人网站，所有内容不会用来进行牟利，也不会利用任何形式的广告来间接获益，纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。