825. Friends Of Appropriate Ages - Medium

标签：requests   The   自己的   reference   ota   -o   自己   tps   lse   

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] + 7
• age[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.

考虑到数据规模，先用hashmap存ages中的年龄和出现次数。checkRequest(a, b)检查年龄a是否会对年龄b发起friend request。遍历ages，如果年龄a会对年龄b发起request，如果a != b，这一对pair产生的request总数= map[a] * map[b]；如果a = b，这一对pair产生的request总数要去除自己对自己的friend request，即map[a] * (map[a] - 1)

注意：check的顺序，应该先check是否会request再计算数量，年龄相同的两个人不一定会相互request

time: O(N^2), space: O(N)

class Solution {
    public int numFriendRequests(int[] ages) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int i : ages) {
            map.put(i, map.getOrDefault(i, 0) + 1);
        }

        int cnt = 0;
        for(Integer i : map.keySet()) {
            for(Integer j : map.keySet()) {
                if(checkRequest(i, j)) {
                    if(i != j)
                        cnt += map.get(i) * map.get(j);
                    else
                        cnt += map.get(i) * (map.get(i) - 1);
                }     
            }
        }
        return cnt;
    }
    
    private boolean checkRequest(int A, int B) {
        if(B <= 0.5 * A + 7) return false;
        if(B > A) return false;
        if(B > 100 && A < 100) return false;
        return true;
    }
}

reference: https://leetcode.com/problems/friends-of-appropriate-ages/discuss/127029/C++JavaPython-Easy-and-Straight-Forward

825. Friends Of Appropriate Ages - Medium

标签：requests   The   自己的   reference   ota   -o   自己   tps   lse   

原文地址：https://www.cnblogs.com/fatttcat/p/10090357.html