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【LeetCode】1013. Pairs of Songs With Total Durations Divisible by 60 解题报告(Python)


id: fuxuemingzhu

个人博客: ​​​http://fuxuemingzhu.cn/​​​


目录


  • ​​题目描述​​
  • ​​题目大意​​
  • ​​解题方法​​
  • ​​日期​​


题目地址:​​https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/​​

题目描述

In a list of songs, the ​​i​​​-th song has a duration of ​​time[i]​​ seconds.

Return the number of pairs of songs for which their total duration in seconds is divisible by ​​60​​​. Formally, we want the number of indices ​​i < j​​​ with ​​(time[i] + time[j]) % 60 == 0​​.

Example 1:

Input: [30,20,150,100,40]
Output: 3
Explanation: Three pairs have a total duration divisible by 60:
(time[0] = 30, time[2] = 150): total duration 180
(time[1] = 20, time[3] = 100): total duration 120
(time[1] = 20, time[4] = 40): total duration 60

Example 2:

Input: [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.

Note:


  1. ​1 <= time.length <= 60000​
  2. ​1 <= time[i] <= 500​

题目大意

统计有多少两个数,​​i < j​​​并且​​(time[i] + time[j]) % 60 == 0​​。相当于two sum的改进。

解题方法

这个题怎么想?注意看Note给的提示!首先给出的数组长度是60000,那么是复杂度应该是O(N),看到每个元素的大小是1~500,那么很容易想到,是不是可以对time的元素进行统计,然后统计每两个数字的和是不是60的倍数即可。

首先需要统计每个数字出现的次数。

如果两个数字不同的话,并且这两个数字的和是60的倍数,直接把两个数字的个数相乘。

当然第二个例子提醒了我们,可以在同样的数字中选择两个,所以从相同的数字中任意选择两个,公式是 N * (N - 1) / 2。

Python代码如下:

class Solution(object):
def numPairsDivisibleBy60(self, time):
"""
:type time: List[int]
:rtype: int
"""
count = collections.Counter(time)
key = list(count.keys())
N = len(key)
res = 0
for i, t in enumerate(key):
for j in range(i, N):
if (t + key[j]) % 60 == 0:
if i == j:
res += count[t] * (count[t] - 1) / 2
else:
res += count[t] * count[key[j]]
return res

日期

2019 年 3 月 21 日 —— 好久不刷题,重拾有点难



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