1010 Pairs of Songs With Total Durations Divisible by 60 总持续时间可被 60 整除的歌曲
Description:
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:
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
题目描述:
在歌曲列表中，第 i 首歌曲的持续时间为 time[i] 秒。
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 :
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 <= time.length <= 60000
1 <= time[i] <= 500
思路:
- 最直接的思路, 用两个循环遍历查找和为 60的倍数的情况
时间复杂度O(n ^ 2), 空间复杂度O(1) - 参考LeetCode #1 Two Sum 两数之和, 用一个数组记录已经出现过的 time数组中的值, 找到匹配值即可, 思想类似双指针
时间复杂度O(n), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int numPairsDivisibleBy60(vector<int>& time)
{
int result = 0, count[60]{0};
for (auto t : time)
{
result += count[(60 - t % 60) % 60];
count[t % 60]++;
}
return result;
}
};
Java:
class Solution {
public int numPairsDivisibleBy60(int[] time) {
int result = 0, count[] = new int[60];
for (int t : time) {
result += count[(60 - t % 60) % 60];
count[t % 60]++;
}
return result;
}
}
Python:
class Solution:
def numPairsDivisibleBy60(self, time: List[int]) -> int:
count, result = [0] * 60, 0
for t in time:
result += count[(60 - t % 60) % 60]
count[t % 60] += 1
return result
