标签：put ota wing not ges app length iat ati
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] < 100Otherwise, 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 }
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:
1 class Solution { 2 public int numFriendRequests(int[] ages) { 3 if(ages == null || ages.length == 0){ 4 return 0; 5 } 6 7 int [] count = new int[121]; 8 for(int age : ages){ 9 count[age]++; 10 } 11 12 int [] sum = new int[121]; 13 for(int i = 1; i<121; i++){ 14 sum[i] = sum[i-1] + count[i]; 15 } 16 17 int res = 0; 18 for(int i = 15; i<121; i++){ 19 if(count[i] == 0){ 20 continue; 21 } 22 23 int bCount = sum[i] - sum[i/2+7]; 24 res += bCount * count[i] - count[i]; 25 } 26 27 return res; 28 } 29 }
LeetCode 825. Friends Of Appropriate Ages
标签：put ota wing not ges app length iat ati
原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12015827.html
