【python3】leetcode 825. Friends Of Appropriate Ages (Medium)

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本文深入探讨了适龄交友算法的实现，通过分析年龄限制条件，提出了一种高效计算总交友请求数量的方法，避免了传统遍历方式的时间复杂度过高的问题。

825. Friends Of Appropriate Ages (Medium)

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.

这道题如果遍历a，b，时间复杂度是O(n2)，妥妥超时

但是由于年龄《120，所以遍历年龄的count，

因为一对a->b满足条件的话，所有a年龄的人可以对所有b年龄的人发起朋友请求，总共数量count(a) * count(b)

而如果ab年龄相同，由于不能对自己发起朋友请求，所以总共数量count(a) * ( count(a) - 1)

class Solution(object):
    def numFriendRequests(self, ages):
        """
        :type ages: List[int]
        :rtype: int
        """
        count = collections.Counter(ages)
        num = 0
        ageset = list(set(ages))
        for i in range(len(ageset)):
            for j in range(len(ageset)):
                if not (ageset[i] <= 0.5*ageset[j] + 7 or (ageset[i] > ageset[j]) or (ageset[i] > 100 and ageset[j] < 100)):
                    if ageset[i] == ageset[j]:num += count[ageset[i]]*(count[ageset[i]]-1)
                    else:num += count[ageset[i]]*count[ageset[j]]
        return num