【LeetCode】1013. Pairs of Songs With Total Durations Divisible by 60(型取り)
1525 ワード
【LeetCode】1013. Pairs of Songs With Total Durations Divisible by 60(型取り)
タイトル
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 i < j iExample 1:
Example 2:
Note:
構想
この問題は実はとても簡単で、注意するところは2つあります:(1)(a+b)%60=(a%60+b%60)%60(a+b)%60=(a%60+b%60)%60(a+b)%60=(a%60+b%60)%60.従って、所定の配列を巡回する際に、各数を直接60に型取ることができる.(2)データを離散化する.
コード#コード#
タイトル
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 i < j i
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
構想
この問題は実はとても簡単で、注意するところは2つあります:(1)(a+b)%60=(a%60+b%60)%60(a+b)%60=(a%60+b%60)%60(a+b)%60=(a%60+b%60)%60.従って、所定の配列を巡回する際に、各数を直接60に型取ることができる.(2)データを離散化する.
コード#コード#
class Solution {
public:
int numPairsDivisibleBy60(vector& time) {
int i, j, k, s, ans;
mapm;
m.clear();
ans = 0;
for(i=0; i