Add solution and test-cases for problem 3693

leetcode/3601-3700/3693.Climbing-Stairs-II/README.md

@@ -1,28 +1,62 @@
 # [3693.Climbing Stairs II][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+You are climbing a staircase with `n + 1` steps, numbered from 0 to `n`.
+
+You are also given a **1-indexed** integer array `costs` of length `n`, where `costs[i]` is the cost of step `i`.
+
+From step `i`, you can jump **only** to step `i + 1`, `i + 2`, or `i + 3`. The cost of jumping from step `i` to step `j` is defined as: `costs[j] + (j - i)^2`
+
+You start from step 0 with `cost = 0`.
+
+Return the **minimum** total cost to reach step `n`.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: n = 4, costs = [1,2,3,4]
+
+Output: 13
+
+Explanation:
+
+One optimal path is 0 → 1 → 2 → 4
+
+Jump	Cost Calculation	Cost
+0 → 1	costs[1] + (1 - 0)2 = 1 + 1	2
+1 → 2	costs[2] + (2 - 1)2 = 2 + 1	3
+2 → 4	costs[4] + (4 - 2)2 = 4 + 4	8
+Thus, the minimum total cost is 2 + 3 + 8 = 13
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
+```
+Input: n = 4, costs = [5,1,6,2]
 
-### 思路1
-> ...
-Climbing Stairs II
-```go
+Output: 11
+
+Explanation:
+
+One optimal path is 0 → 2 → 4
+
+Jump	Cost Calculation	Cost
+0 → 2	costs[2] + (2 - 0)2 = 1 + 4	5
+2 → 4	costs[4] + (4 - 2)2 = 2 + 4	6
+Thus, the minimum total cost is 5 + 6 = 11
 ```
 
+**Example 3:**
+
+```
+Input: n = 3, costs = [9,8,3]
+
+Output: 12
+
+Explanation:
+
+The optimal path is 0 → 3 with total cost = costs[3] + (3 - 0)2 = 3 + 9 = 12
+```
 
 ## 结语
 

leetcode/3601-3700/3693.Climbing-Stairs-II/Solution.go

@@ -1,5 +1,22 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+const INF = 0x7fffffff
+
+func Solution(n int, costs []int) int {
+	dp := make([]int, n+1)
+	for i := 1; i <= n; i++ {
+		dp[i] = INF
+	}
+	for i := 1; i <= n; i++ {
+		if i >= 1 {
+			dp[i] = min(dp[i], dp[i-1]+costs[i-1]+1)
+		}
+		if i >= 2 {
+			dp[i] = min(dp[i], dp[i-2]+costs[i-1]+4)
+		}
+		if i >= 3 {
+			dp[i] = min(dp[i], dp[i-3]+costs[i-1]+9)
+		}
+	}
+	return dp[n]
 }

leetcode/3601-3700/3693.Climbing-Stairs-II/Solution_test.go

@@ -10,30 +10,31 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		n      int
+		costs  []int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", 4, []int{1, 2, 3, 4}, 13},
+		{"TestCase2", 4, []int{5, 1, 6, 2}, 11},
+		{"TestCase3", 3, []int{9, 8, 3}, 12},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.n, c.costs)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
+					c.expect, got, c.n, c.costs)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }