LeetCode 825. Friends Of Appropriate Ages
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:
class Solution {
public int numFriendRequests(int[] ages) {
if(ages == null || ages.length == 0){
return 0;
}
HashMap<Integer, Integer> hm = new HashMap<>();
for(int age : ages){
hm.put(age, hm.getOrDefault(age, 0) + 1);
}
int res = 0;
for(int a : hm.keySet()){
for(int b : hm.keySet()){
if(couldSendRequest(a, b)){
res += hm.get(a) * (hm.get(b) - (a == b ? 1 : 0));
}
}
}
return res;
}
private boolean couldSendRequest(int a, int b){
return !(b <= a*0.5 + 7 || b > a || (b > 100 && a < 100));
}
}
With 3 conditions, we only care the count of B in range (a/2+7, a].
Get the sum count of b and * a count - a count since people can't sent friend request to themselves.
Since A > B >= 0.5*A+7, A > 0.5*A+7. Then A>14. Thus i is started from 15.
Time Complexity: O(n).
Space: O(1).
AC Java:
class Solution {
public int numFriendRequests(int[] ages) {
if(ages == null || ages.length == 0){
return 0;
}
int [] count = new int[121];
for(int age : ages){
count[age]++;
}
int [] sum = new int[121];
for(int i = 1; i<121; i++){
sum[i] = sum[i-1] + count[i];
}
int res = 0;
for(int i = 15; i<121; i++){
if(count[i] == 0){
continue;
}
int bCount = sum[i] - sum[i/2+7];
res += bCount * count[i] - count[i];
}
return res;
}
}
最新文章
- Reveal UI 分析工具分析手机 App
- 收缩TempDB的办法（转载）
- Java Web Service 学习
- Linux_服务
- STP详解
- 京东拍拍网 笔试 搞java的去考C++ 苦逼
- Head First设计模式-观察者模式
- （ASP.net）利用Application对象制作简单聊天室
- Double跟double
- asp.net学习之 数据绑定控件--List数据绑定控件
- KVM克隆 快照
- 数据结构 之 并查集（Disjoint Set）
- Django 的认识，面试题
- 使用JfreeChart生成图表遇到的问题
- hdu 4486 Pen Counts
- 289. Game of Life数组生存游戏
- 自学Linux Shell3.6-文件查看命令file cat more less tail head
- Ubuntu命令行
- [Unity算法]斜抛运动
- debug_toolbar工作原理