Description
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
Input: [16,16]
Output: 2
Explanation: 2 people friend request each other.
Note
1 <= ages.length <= 20000.
1 <= ages[i] <= 120.
Code
Wrong Solution
class Solution(object):
def numFriendRequests(self, ages):
"""
:type ages: List[int]
:rtype: int
"""
N = len(ages)
nums = [0]*121
for i in ages:
nums[i] += 1
dp = [0]*121
for i in range(1, 121):
dp[i] = dp[i-1] + nums[i]
ret = 0
for i in range(120, 0, -1):
if nums[i] == 0:
continue
ret += nums[i]*(nums[i]-1) // 在过滤条件前修改数据，我的思维习惯不是很干净
th = i//2 + 7
if th >= i:
continue
ret += nums[i]* (dp[i-1] - dp[th])
return ret
AC Solution
class Solution(object):
def numFriendRequests(self, ages):
"""
:type ages: List[int]
:rtype: int
"""
N = len(ages)
nums = [0]*121
for i in ages:
nums[i] += 1
dp = [0]*121
for i in range(1, 121):
dp[i] = dp[i-1] + nums[i]
ret = 0
for i in range(120, 0, -1):
if nums[i] == 0:
continue
th = i//2 + 7
if th >= i:
continue
ret += nums[i]*(nums[i]-1)
ret += nums[i]* (dp[i-1] - dp[th])
return ret
总结
犯了思维不干净的错误的题目
1524. Number of Sub-arrays With Odd Sum
