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feat: add solutions to lc problem: No.0651
No.0651.4 Keys Keyboard
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‎solution/0600-0699/0651.4 Keys Keyboard/README.md

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Original file line numberDiff line numberDiff line change
@@ -51,22 +51,94 @@ A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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定义 $dp[i]$ 表示前 $i$ 个按键可以显示的最大个数。
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我们可以发现，要显示最多的 `A`，要么一直按 `A`，要么以 `Ctrl-V` 结束。
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- 一直按 `A` 的情况，满足 $dp[i] = i$。
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- 以 `Ctrl-V` 结束的情况，我们枚举对应的 `Ctrl-A` 的位置 $j$，可以得到 $dp[i]=max(dp[i], dp[j-1] \times (i - j))$。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maxA(self, n: int) -> int:
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dp = list(range(n + 1))
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for i in range(3, n + 1):
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for j in range(2, i - 1):
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dp[i] = max(dp[i], dp[j - 1] * (i - j))
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return dp[-1]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maxA(int n) {
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int[] dp = new int[n + 1];
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for (int i = 0; i < n + 1; ++i) {
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dp[i] = i;
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}
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for (int i = 3; i < n + 1; ++i) {
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for (int j = 2; j < i - 1; ++j) {
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dp[i] = Math.max(dp[i], dp[j - 1] * (i - j));
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}
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}
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return dp[n];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxA(int n) {
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vector<int> dp(n + 1);
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iota(dp.begin(), dp.end(), 0);
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for (int i = 3; i < n + 1; ++i) {
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for (int j = 2; j < i - 1; ++j) {
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dp[i] = max(dp[i], dp[j - 1] * (i - j));
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}
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}
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return dp[n];
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}
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};
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```
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### **Go**
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```go
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func maxA(n int) int {
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dp := make([]int, n+1)
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for i := range dp {
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dp[i] = i
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}
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for i := 3; i < n+1; i++ {
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for j := 2; j < i-1; j++ {
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dp[i] = max(dp[i], dp[j-1]*(i-j))
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}
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}
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return dp[n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0600-0699/0651.4 Keys Keyboard/README_EN.md

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@@ -48,13 +48,74 @@ A, A, A, Ctrl A, Ctrl C, Ctrl V, Ctrl V
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### **Python3**
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```python
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class Solution:
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def maxA(self, n: int) -> int:
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dp = list(range(n + 1))
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for i in range(3, n + 1):
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for j in range(2, i - 1):
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dp[i] = max(dp[i], dp[j - 1] * (i - j))
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return dp[-1]
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```
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### **Java**
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```java
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class Solution {
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public int maxA(int n) {
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int[] dp = new int[n + 1];
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for (int i = 0; i < n + 1; ++i) {
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dp[i] = i;
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}
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for (int i = 3; i < n + 1; ++i) {
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for (int j = 2; j < i - 1; ++j) {
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dp[i] = Math.max(dp[i], dp[j - 1] * (i - j));
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}
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}
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return dp[n];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maxA(int n) {
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vector<int> dp(n + 1);
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iota(dp.begin(), dp.end(), 0);
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for (int i = 3; i < n + 1; ++i) {
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for (int j = 2; j < i - 1; ++j) {
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dp[i] = max(dp[i], dp[j - 1] * (i - j));
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}
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}
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return dp[n];
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}
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};
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```
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### **Go**
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```go
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func maxA(n int) int {
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dp := make([]int, n+1)
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for i := range dp {
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dp[i] = i
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}
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for i := 3; i < n+1; i++ {
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for j := 2; j < i-1; j++ {
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dp[i] = max(dp[i], dp[j-1]*(i-j))
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}
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}
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return dp[n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/0600-0699/0651.4 Keys Keyboard/Solution.cpp

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class Solution {
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public:
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int maxA(int n) {
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vector<int> dp(n + 1);
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iota(dp.begin(), dp.end(), 0);
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for (int i = 3; i < n + 1; ++i) {
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for (int j = 2; j < i - 1; ++j) {
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dp[i] = max(dp[i], dp[j - 1] * (i - j));
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}
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}
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return dp[n];
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}
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};

‎solution/0600-0699/0651.4 Keys Keyboard/Solution.go

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func maxA(n int) int {
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dp := make([]int, n+1)
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for i := range dp {
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dp[i] = i
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}
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for i := 3; i < n+1; i++ {
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for j := 2; j < i-1; j++ {
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dp[i] = max(dp[i], dp[j-1]*(i-j))
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}
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}
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return dp[n]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/0600-0699/0651.4 Keys Keyboard/Solution.java

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class Solution {
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public int maxA(int n) {
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int[] dp = new int[n + 1];
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for (int i = 0; i < n + 1; ++i) {
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dp[i] = i;
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}
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for (int i = 3; i < n + 1; ++i) {
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for (int j = 2; j < i - 1; ++j) {
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dp[i] = Math.max(dp[i], dp[j - 1] * (i - j));
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}
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}
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return dp[n];
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}
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}

‎solution/0600-0699/0651.4 Keys Keyboard/Solution.py

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class Solution:
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def maxA(self, n: int) -> int:
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dp = list(range(n + 1))
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for i in range(3, n + 1):
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for j in range(2, i - 1):
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dp[i] = max(dp[i], dp[j - 1] * (i - j))
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return dp[-1]

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