[Swift Weekly Contest 128]LeetCode1013. 总持续时间可被 60 整除的歌曲 | Pairs of Songs With Total Durations Divisible by 60

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

返回其总持续时间（以秒为单位）可被 60 整除的歌曲对的数量。形式上，我们希望索引的数字  i < j 且有 (time[i] + time[j]) % 60 == 0。

示例 1：

输入：[30,20,150,100,40]
输出：3
解释：这三对的总持续时间可被 60 整数：
(time[0] = 30, time[2] = 150): 总持续时间 180
(time[1] = 20, time[3] = 100): 总持续时间 120
(time[1] = 20, time[4] = 40): 总持续时间 60

示例 2：

输入：[60,60,60]
输出：3
解释：所有三对的总持续时间都是 120，可以被 60 整数。 

提示：

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

Memory Usage: 19.4 MB

 1 class Solution {
 2     func numPairsDivisibleBy60(_ time: [Int]) -> Int {
 3         var n:Int = time.count
 4         var cnt:[Int] = [Int](repeating:0,count:60)
 5         var ans:Int = 0
 6         for i in 0..<n
 7         {
 8             var t:Int = time[i] % 60
 9             ans += cnt[(60 - t) % 60]
10             cnt[time[i] % 60] += 1
11         }
12         return ans
13
14     }
15 }

原文地址：https://www.cnblogs.com/strengthen/p/10546217.html