Merge pull request #849 from 0xff-dev/650

Add solution and test-cases for problem 650

Author: 6boris

leetcode/601-700/0650.2-Keys-Keyboard/README.md

@@ -1,28 +1,30 @@
 # [650.2 Keys Keyboard][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
+There is only one character `'A'` on the screen of a notepad. You can perform one of two operations on this notepad for each step:
+
+- Copy All: You can copy all the characters present on the screen (a partial copy is not allowed).
+- Paste: You can paste the characters which are copied last time.
+
+Given an integer `n`, return the minimum number of operations to get the character `'A'` exactly `n` times on the screen.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: n = 3
+Output: 3
+Explanation: Initially, we have one character 'A'.
+In step 1, we use Copy All operation.
+In step 2, we use Paste operation to get 'AA'.
+In step 3, we use Paste operation to get 'AAA'.
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-2 Keys Keyboard
-```go
 ```
-
+Input: n = 1
+Output: 0
+```
 
 ## 结语
 

leetcode/601-700/0650.2-Keys-Keyboard/Solution.go

@@ -1,5 +1,55 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(n int) int {
+	if n == 1 {
+		return 0
+	}
+	// 存在一个当前复制几个问题
+	// 1  复制了1次
+	// dp[i][j] 表示通过复制j个得到i个元素需要的最少次数，0就是无法到达
+	dp := make([][]int, n+1)
+	for i := 0; i <= n; i++ {
+		dp[i] = make([]int, n+1)
+		if i > 1 {
+			for j := 0; j <= n; j++ {
+				dp[i][j] = -1
+			}
+		}
+	}
+	for i := 2; i <= n; i++ {
+		// 通过粘贴nci 到达i
+		for ct := i - 1; ct > 0; ct-- {
+			start := i - ct // 从那个位置开始
+			if start > ct && dp[start][ct] != -1 {
+				if r := dp[start][ct] + 1; dp[i][ct] == -1 || dp[i][ct] > r {
+					dp[i][ct] = r
+				}
+			}
+			if start == ct {
+				op := -1
+				if start == 1 {
+					op = 0
+				}
+				for pn := 1; pn < start; pn++ {
+					if dp[start][pn] != -1 && (op == -1 || dp[start][pn] < op) {
+						op = dp[start][pn]
+					}
+				}
+				op += 2 //copy and past
+				if dp[i][ct] == -1 || dp[i][ct] > op {
+					dp[i][ct] = op
+				}
+			}
+		}
+	}
+	op := -1
+	for i := 1; i < n; i++ {
+		if dp[n][i] == -1 {
+			continue
+		}
+		if op == -1 || dp[n][i] < op {
+			op = dp[n][i]
+		}
+	}
+	return op
 }

leetcode/601-700/0650.2-Keys-Keyboard/Solution_test.go

@@ -10,12 +10,12 @@ 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},
+		{"TestCase1", 3, 3},
+		{"TestCase2", 1, 0},
+		{"TestCase3", 99, 17},
 	}
 
 	//	开始测试
@@ -30,10 +30,10 @@ func TestSolution(t *testing.T) {
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }