题目如下:
In a list of songs, the
i
-th song has a duration oftime[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 indicesi < 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 <= time.length <= 60000
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