✨ Added solution to 1010

Sathish Babu · Sathish Babu · commit 59518b841fcf · 2020-06-03T08:04:39.000+05:30
diff --git a/src/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/README.md b/src/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/README.md
@@ -5,11 +5,27 @@
 
 ## 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 such that i < j with (time[i] + time[j]) % 60 == 0.
+
+**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 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: [60,60,60]
+Output: 3
+Explanation: All three pairs have a total duration of 120, which is divisible by 60.
 ```
 
 ## 题意
diff --git a/src/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/Solution.go b/src/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/Solution.go
@@ -1,5 +1,15 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(time []int) int {
+	ans, hash := 0, make([]int, 60)
+	for _, t := range time {
+		v := t % 60
+		if v == 0 {
+			ans += hash[0]
+		} else {
+			ans += hash[60-v]
+		}
+		hash[v]++
+	}
+	return ans
 }
diff --git a/src/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/Solution_test.go b/src/1010.Pairs-of-Songs-With-Total-Durations-Divisible-by-60/Solution_test.go
@@ -10,12 +10,11 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		inputs []int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase", []int{30, 20, 150, 100, 40}, 3},
+		{"TestCase", []int{60, 60, 60}, 3},
 	}
 
 	//	开始测试
