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feat: add solutions to lcp problem: No.19 (#1346)
LCP 19. 秋叶收藏集
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‎lcp/LCP 19. 秋叶收藏集/README.md

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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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我们定义 $f[i][j]$ 表示对于第 $i$ 片叶子，处于状态 $j$ 时的最小操作次数，其中 $j$ 表示：
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- 状态 $0$ 表示第 $i$ 片叶子处于左边的红色部分；
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- 状态 $1$ 表示第 $i$ 片叶子处于中间的黄色部分；
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- 状态 $2$ 表示第 $i$ 片叶子处于右边的红色部分。
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初始时，如果第 $0$ 片叶子为红色，那么 $f[0][0] = 0$，如果第 $0$ 片叶子为黄色，那么 $f[0][0] = 1$。对于其余的状态，我们初始化为 $+\infty$，表示在这种状态下无法完成任务。
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考虑 $f[i][j]$，其中 $i \ge 1$：
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如果第 $i$ 片叶子为红色，那么 $f[i][0]$ 只能从 $f[i-1][0]$ 转移而来，而 $f[i][1]$ 可以从 $f[i-1][0]$ 和 $f[i-1][1]$ 转移而来，此时需要额外操作一次，而 $f[i][2]$ 可以从 $f[i-1][1]$ 和 $f[i-1][2]$ 转移而来，此时不需要额外操作一次。
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如果第 $i$ 片叶子为黄色，那么 $f[i][0]$ 只能从 $f[i-1][0]$ 转移而来，并且需要额外操作一次，而 $f[i][1]$ 可以从 $f[i-1][0]$ 和 $f[i-1][1]$ 转移而来，此时不需要额外操作一次，而 $f[i][2]$ 可以从 $f[i-1][1]$ 和 $f[i-1][2]$ 转移而来，并且需要额外操作一次。
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最终的答案即为 $f[n-1][2]$，其中 $n$ 表示红叶和黄叶的总数。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组的长度。
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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 minimumOperations(self, leaves: str) -> int:
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n = len(leaves)
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f = [[inf] * 3 for _ in range(n)]
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f[0][0] = int(leaves[0] == "y")
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for i in range(1, n):
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if leaves[i] == "r":
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f[i][0] = f[i - 1][0]
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f[i][1] = min(f[i - 1][0], f[i - 1][1]) + 1
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f[i][2] = min(f[i - 1][2], f[i - 1][1])
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else:
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f[i][0] = f[i - 1][0] + 1
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f[i][1] = min(f[i - 1][0], f[i - 1][1])
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f[i][2] = min(f[i - 1][2], f[i - 1][1]) + 1
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return f[n - 1][2]
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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 minimumOperations(String leaves) {
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int n = leaves.length();
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final int inf = 1 << 30;
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var f = new int[n][3];
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for (var g : f) {
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Arrays.fill(g, inf);
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}
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f[0][0] = leaves.charAt(0) == 'r' ? 0 : 1;
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for (int i = 1; i < n; ++i) {
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if (leaves.charAt(i) == 'r') {
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f[i][0] = f[i - 1][0];
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]) + 1;
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]);
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} else {
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f[i][0] = f[i - 1][0] + 1;
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]);
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]) + 1;
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}
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}
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return f[n - 1][2];
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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 minimumOperations(string leaves) {
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int n = leaves.size();
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int f[n][3];
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memset(f, 0x3f, sizeof(f));
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f[0][0] = leaves[0] == 'y';
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for (int i = 1; i < n; ++i) {
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if (leaves[i] == 'r') {
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f[i][0] = f[i - 1][0];
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f[i][1] = min(f[i - 1][0], f[i - 1][1]) + 1;
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f[i][2] = min(f[i - 1][2], f[i - 1][1]);
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} else {
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f[i][0] = f[i - 1][0] + 1;
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f[i][1] = min(f[i - 1][0], f[i - 1][1]);
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f[i][2] = min(f[i - 1][2], f[i - 1][1]) + 1;
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}
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}
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return f[n - 1][2];
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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 minimumOperations(leaves string) int {
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n := len(leaves)
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f := make([][3]int, n)
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inf := 1 << 30
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for i := range f {
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f[i] = [3]int{inf, inf, inf}
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}
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if leaves[0] == 'y' {
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f[0][0] = 1
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} else {
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f[0][0] = 0
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}
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for i := 1; i < n; i++ {
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if leaves[i] == 'r' {
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f[i][0] = f[i-1][0]
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f[i][1] = min(f[i-1][0], f[i-1][1]) + 1
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f[i][2] = min(f[i-1][2], f[i-1][1])
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} else {
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f[i][0] = f[i-1][0] + 1
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f[i][1] = min(f[i-1][0], f[i-1][1])
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f[i][2] = min(f[i-1][2], f[i-1][1]) + 1
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}
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}
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return f[n-1][2]
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}
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func min(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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### **TypeScript**
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```ts
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function minimumOperations(leaves: string): number {
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const n = leaves.length;
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const inf = 1 << 30;
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const f: number[][] = new Array(n)
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.fill(0)
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.map(() => new Array(3).fill(inf));
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f[0][0] = leaves[0] === 'y' ? 1 : 0;
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for (let i = 1; i < n; ++i) {
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if (leaves[i] === 'r') {
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f[i][0] = f[i - 1][0];
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]) + 1;
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]);
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} else {
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f[i][0] = f[i - 1][0] + 1;
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]);
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]) + 1;
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}
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}
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return f[n - 1][2];
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}
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```
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### **...**

‎lcp/LCP 19. 秋叶收藏集/Solution.cpp

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class Solution {
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public:
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int minimumOperations(string leaves) {
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int n = leaves.size();
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int f[n][3];
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memset(f, 0x3f, sizeof(f));
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f[0][0] = leaves[0] == 'y';
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for (int i = 1; i < n; ++i) {
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if (leaves[i] == 'r') {
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f[i][0] = f[i - 1][0];
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f[i][1] = min(f[i - 1][0], f[i - 1][1]) + 1;
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f[i][2] = min(f[i - 1][2], f[i - 1][1]);
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} else {
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f[i][0] = f[i - 1][0] + 1;
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f[i][1] = min(f[i - 1][0], f[i - 1][1]);
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f[i][2] = min(f[i - 1][2], f[i - 1][1]) + 1;
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}
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}
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return f[n - 1][2];
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}
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};

‎lcp/LCP 19. 秋叶收藏集/Solution.go

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func minimumOperations(leaves string) int {
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n := len(leaves)
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f := make([][3]int, n)
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inf := 1 << 30
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for i := range f {
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f[i] = [3]int{inf, inf, inf}
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}
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if leaves[0] == 'y' {
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f[0][0] = 1
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} else {
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f[0][0] = 0
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}
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for i := 1; i < n; i++ {
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if leaves[i] == 'r' {
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f[i][0] = f[i-1][0]
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f[i][1] = min(f[i-1][0], f[i-1][1]) + 1
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f[i][2] = min(f[i-1][2], f[i-1][1])
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} else {
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f[i][0] = f[i-1][0] + 1
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f[i][1] = min(f[i-1][0], f[i-1][1])
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f[i][2] = min(f[i-1][2], f[i-1][1]) + 1
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}
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}
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return f[n-1][2]
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}
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func min(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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}

‎lcp/LCP 19. 秋叶收藏集/Solution.java

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class Solution {
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public int minimumOperations(String leaves) {
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int n = leaves.length();
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final int inf = 1 << 30;
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var f = new int[n][3];
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for (var g : f) {
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Arrays.fill(g, inf);
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}
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f[0][0] = leaves.charAt(0) == 'r' ? 0 : 1;
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for (int i = 1; i < n; ++i) {
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if (leaves.charAt(i) == 'r') {
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f[i][0] = f[i - 1][0];
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]) + 1;
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]);
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} else {
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f[i][0] = f[i - 1][0] + 1;
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]);
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]) + 1;
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}
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}
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return f[n - 1][2];
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}
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}

‎lcp/LCP 19. 秋叶收藏集/Solution.py

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class Solution:
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def minimumOperations(self, leaves: str) -> int:
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n = len(leaves)
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f = [[inf] * 3 for _ in range(n)]
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f[0][0] = int(leaves[0] == "y")
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for i in range(1, n):
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if leaves[i] == "r":
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f[i][0] = f[i - 1][0]
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f[i][1] = min(f[i - 1][0], f[i - 1][1]) + 1
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f[i][2] = min(f[i - 1][2], f[i - 1][1])
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else:
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f[i][0] = f[i - 1][0] + 1
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f[i][1] = min(f[i - 1][0], f[i - 1][1])
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f[i][2] = min(f[i - 1][2], f[i - 1][1]) + 1
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return f[n - 1][2]

‎lcp/LCP 19. 秋叶收藏集/Solution.ts

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function minimumOperations(leaves: string): number {
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const n = leaves.length;
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const inf = 1 << 30;
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const f: number[][] = new Array(n)
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.fill(0)
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.map(() => new Array(3).fill(inf));
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f[0][0] = leaves[0] === 'y' ? 1 : 0;
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for (let i = 1; i < n; ++i) {
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if (leaves[i] === 'r') {
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f[i][0] = f[i - 1][0];
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]) + 1;
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]);
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} else {
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f[i][0] = f[i - 1][0] + 1;
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f[i][1] = Math.min(f[i - 1][0], f[i - 1][1]);
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f[i][2] = Math.min(f[i - 1][2], f[i - 1][1]) + 1;
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}
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}
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return f[n - 1][2];
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}

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