leetcode学习笔记1010
开始之前
问题
Pairs of Songs With Total Durations Divisible by 60
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
方法1
0 和 30 单独计算, 其他的值直接去找对应的值是否存在就可以.
时间复杂度O(n).
空间复杂度O(1).
class Solution {
public int numPairsDivisibleBy60(int[] time) {
int[] count = new int[60];
// 对60取余
for(int t : time)
count[t%60]++;
int res = getPairCouns(count[0]) + getPairCouns(count[30]);
for(int i = 1; i < 30; i++)
res += count[i] * count[60 - i];
return res;
}
private int getPairCouns(int n){
return n*(n-1)/2;
}
}