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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,
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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