Add solution 1010

leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60.go

@@ -0,0 +1,16 @@
+package leetcode
+
+func numPairsDivisibleBy60(time []int) int {
+	counts := make([]int, 60)
+	for _, v := range time {
+		v %= 60
+		counts[v]++
+	}
+	res := 0
+	for i := 1; i < len(counts)/2; i++ {
+		res += counts[i] * counts[60-i]
+	}
+	res += (counts[0] * (counts[0] - 1)) / 2
+	res += (counts[30] * (counts[30] - 1)) / 2
+	return res
+}

leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/1010. Pairs of Songs With Total Durations Divisible by 60_test.go

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+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1010 struct {
+	para1010
+	ans1010
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1010 struct {
+	time []int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1010 struct {
+	one int
+}
+
+func Test_Problem1010(t *testing.T) {
+
+	qs := []question1010{
+
+		{
+			para1010{[]int{30, 20, 150, 100, 40}},
+			ans1010{3},
+		},
+
+		{
+			para1010{[]int{60, 60, 60}},
+			ans1010{3},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1010------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1010, q.para1010
+		fmt.Printf("【input】:%v       【output】:%v\n", p, numPairsDivisibleBy60(p.time))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/README.md

@@ -0,0 +1,63 @@
+# [1010. Pairs of Songs With Total Durations Divisible by 60](https://leetcode.com/problems/pairs-of-songs-with-total-durations-divisible-by-60/)
+
+
+## 题目
+
+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
+
+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，我们希望下标数字 i 和 j 满足  i < j 且有 (time[i] + time[j]) % 60 == 0。
+
+## 解题思路
+
+- 简单题。先将数组每个元素对 60 取余，将它们都转换到 [0,59] 之间。然后在数组中找两两元素之和等于 60 的数对。可以在 0-30 之内对半查找符合条件的数对。对 0 和 30 单独计算。因为多个 0 相加，余数还为 0 。2 个 30 相加之和为 60。
+
+## 代码
+
+```go
+func numPairsDivisibleBy60(time []int) int {
+	counts := make([]int, 60)
+	for _, v := range time {
+		v %= 60
+		counts[v]++
+	}
+	res := 0
+	for i := 1; i < len(counts)/2; i++ {
+		res += counts[i] * counts[60-i]
+	}
+	res += (counts[0] * (counts[0] - 1)) / 2
+	res += (counts[30] * (counts[30] - 1)) / 2
+	return res
+}
+```