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

作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

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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