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  • 【leetcode】1013. Pairs of Songs With Total Durations Divisible by 60

    题目如下:

    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

    解题思路:遍历Input并对其中每个元素与60取模,以余数为key值存入字典dic中,字典的value也key值作为余数出现的次数。接下来再遍历一次Input,求出元素与60取模后的余数,再求出60减去余数的差值,字典dic[差值]所对应的值即为这个元素可以与数组中多少个元素的和能被60整除。

    代码如下:

    class Solution(object):
        def numPairsDivisibleBy60(self, time):
            """
            :type time: List[int]
            :rtype: int
            """
            dic = {}
            for i in time:
                v = i % 60
                dic[v] = dic.setdefault(v,0) + 1
            res = 0
            for i in time:
                v = i % 60
                dic[v] -= 1
                key = 60 -v if v != 0 else 0
                if key in dic:
                    res += dic[key]
            return res
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10559788.html
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