You are given a list of songs where the ith 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
such that i < j
with (time[i] + time[j]) % 60 == 0
.Example 1:
Input: time = [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: time = [60,60,60]
Output: 3
Explanation: All three pairs have a total duration of 120, which is divisible by 60.
Constraints:
`1 <= time.length <= 6 104
*
1 <= time[i] <= 500`题目大意:
求数组中两数和能被60整除
解题思路:
两数和的关系第一时间想到two sum。但由于target是60的倍数,并不固定,所以先用公式求所有数的mod,(time[i] + time[j]) % 60 = time[i] % 60 + time[j] % 60, 这样target就是60了
第二个难点是此题求个数并不是像two sum一样求可行性,所以value to index改成value to count
解题步骤:
N/A
注意事项:
- 对所有数对60求mod,map存value到count
- 如果输入是60,取模后为0, 求(60 - time_mod[i])要对60取模,否则60不在map中,因为60 - time_mod[i] = 60.
Python代码:
1 | # (time[i] + time[j]) % 60 = time[i] % 60 + time[j] % 60 |
算法分析:
时间复杂度为O(n)
,空间复杂度O(n)