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feat: add solutions to lc problem: No.0818
No.0818.Race Car
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‎solution/0800-0899/0818.Race Car/README.md

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@@ -61,22 +61,110 @@
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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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设 $dp[i]$ 表示到达位置 $i$ 的最短指令序列的长度。答案为 $dp[target]$。
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对于任意位置 $i$，都有 $2^{k-1} \leq i \lt 2^k$，并且我们可以有三种方式到达位置 $i$：
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- 如果 $i$ 等于 $2^k-1$，那么我们可以直接执行 $k$ 个 `A` 指令到达位置 $i$，此时 $dp[i] = k$；
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- 否则，我们可以先执行 $k$ 个 `A` 指令到达位置 $2^k-1$，然后执行 `R` 指令，剩余距离为 $2^k-1-i$，此时 $dp[i] = dp[2^k-1-i] + k + 1$；我们也可以先执行 $k-1$ 个 `A` 指令到达位置 $2^{k-1}-1$，然后执行 `R` 指令，接着执行 $j$（其中 $0 \le j \lt k$） 个 `A`，再执行 `R`，剩余距离为 $i - 2^{k-1} + 2^j$，此时 $dp[i] = dp[i - 2^{k-1} + 2^j] + k - 1 + j + 2$。求出 $dp[i]$ 的最小值即可。
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时间复杂度 $O(n \log n)$，其中 $n$ 为 $target$。
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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 racecar(self, target: int) -> int:
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dp = [0] * (target + 1)
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for i in range(1, target + 1):
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k = i.bit_length()
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if i == 2**k - 1:
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dp[i] = k
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continue
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dp[i] = dp[2**k - 1 - i] + k + 1
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for j in range(k - 1):
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dp[i] = min(dp[i], dp[i - (2 ** (k - 1) - 2**j)] + k - 1 + j + 2)
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return dp[target]
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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 racecar(int target) {
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int[] dp = new int[target + 1];
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for (int i = 1; i <= target; ++i) {
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int k = 32 - Integer.numberOfLeadingZeros(i);
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if (i == (1 << k) - 1) {
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dp[i] = k;
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continue;
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}
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dp[i] = dp[(1 << k) - 1 - i] + k + 1;
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for (int j = 0; j < k; ++j) {
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dp[i] = Math.min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
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}
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}
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return dp[target];
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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 racecar(int target) {
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vector<int> dp(target + 1);
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for (int i = 1; i <= target; ++i) {
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int k = 32 - __builtin_clz(i);
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if (i == (1 << k) - 1) {
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dp[i] = k;
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continue;
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}
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dp[i] = dp[(1 << k) - 1 - i] + k + 1;
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for (int j = 0; j < k; ++j) {
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dp[i] = min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
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}
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}
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return dp[target];
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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 racecar(target int) int {
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dp := make([]int, target+1)
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for i := 1; i <= target; i++ {
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k := bits.Len(uint(i))
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if i == (1<<k)-1 {
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dp[i] = k
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continue
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}
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dp[i] = dp[(1<<k)-1-i] + k + 1
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for j := 0; j < k; j++ {
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dp[i] = min(dp[i], dp[i-(1<<(k-1))+(1<<j)]+k-1+j+2)
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}
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}
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return dp[target]
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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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### **...**

‎solution/0800-0899/0818.Race Car/README_EN.md

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@@ -60,13 +60,90 @@ Your position goes from 0 --&gt; 1 --&gt; 3 --&gt; 7 --&gt; 7 --&gt; 6.
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### **Python3**
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```python
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class Solution:
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def racecar(self, target: int) -> int:
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dp = [0] * (target + 1)
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for i in range(1, target + 1):
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k = i.bit_length()
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if i == 2**k - 1:
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dp[i] = k
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continue
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dp[i] = dp[2**k - 1 - i] + k + 1
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for j in range(k - 1):
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dp[i] = min(dp[i], dp[i - (2 ** (k - 1) - 2**j)] + k - 1 + j + 2)
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return dp[target]
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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 racecar(int target) {
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int[] dp = new int[target + 1];
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for (int i = 1; i <= target; ++i) {
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int k = 32 - Integer.numberOfLeadingZeros(i);
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if (i == (1 << k) - 1) {
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dp[i] = k;
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continue;
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}
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dp[i] = dp[(1 << k) - 1 - i] + k + 1;
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for (int j = 0; j < k; ++j) {
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dp[i] = Math.min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
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}
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}
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return dp[target];
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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 racecar(int target) {
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vector<int> dp(target + 1);
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for (int i = 1; i <= target; ++i) {
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int k = 32 - __builtin_clz(i);
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if (i == (1 << k) - 1) {
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dp[i] = k;
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continue;
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}
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dp[i] = dp[(1 << k) - 1 - i] + k + 1;
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for (int j = 0; j < k; ++j) {
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dp[i] = min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
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}
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}
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return dp[target];
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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 racecar(target int) int {
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dp := make([]int, target+1)
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for i := 1; i <= target; i++ {
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k := bits.Len(uint(i))
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if i == (1<<k)-1 {
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dp[i] = k
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continue
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}
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dp[i] = dp[(1<<k)-1-i] + k + 1
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for j := 0; j < k; j++ {
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dp[i] = min(dp[i], dp[i-(1<<(k-1))+(1<<j)]+k-1+j+2)
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}
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}
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return dp[target]
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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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### **...**

‎solution/0800-0899/0818.Race Car/Solution.cpp

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class Solution {
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public:
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int racecar(int target) {
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vector<int> dp(target + 1);
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for (int i = 1; i <= target; ++i) {
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int k = 32 - __builtin_clz(i);
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if (i == (1 << k) - 1) {
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dp[i] = k;
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continue;
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}
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dp[i] = dp[(1 << k) - 1 - i] + k + 1;
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for (int j = 0; j < k; ++j) {
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dp[i] = min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
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}
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}
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return dp[target];
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}
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};

‎solution/0800-0899/0818.Race Car/Solution.go

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func racecar(target int) int {
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dp := make([]int, target+1)
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for i := 1; i <= target; i++ {
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k := bits.Len(uint(i))
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if i == (1<<k)-1 {
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dp[i] = k
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continue
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}
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dp[i] = dp[(1<<k)-1-i] + k + 1
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for j := 0; j < k; j++ {
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dp[i] = min(dp[i], dp[i-(1<<(k-1))+(1<<j)]+k-1+j+2)
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}
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}
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return dp[target]
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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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}

‎solution/0800-0899/0818.Race Car/Solution.java

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class Solution {
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public int racecar(int target) {
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int[] dp = new int[target + 1];
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for (int i = 1; i <= target; ++i) {
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int k = 32 - Integer.numberOfLeadingZeros(i);
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if (i == (1 << k) - 1) {
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dp[i] = k;
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continue;
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}
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dp[i] = dp[(1 << k) - 1 - i] + k + 1;
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for (int j = 0; j < k; ++j) {
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dp[i] = Math.min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
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}
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}
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return dp[target];
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}
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}

‎solution/0800-0899/0818.Race Car/Solution.py

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class Solution:
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def racecar(self, target: int) -> int:
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dp = [0] * (target + 1)
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for i in range(1, target + 1):
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k = i.bit_length()
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if i == 2**k - 1:
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dp[i] = k
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continue
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dp[i] = dp[2**k - 1 - i] + k + 1
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for j in range(k - 1):
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dp[i] = min(dp[i], dp[i - (2 ** (k - 1) - 2**j)] + k - 1 + j + 2)
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return dp[target]

‎solution/1100-1199/1147.Longest Chunked Palindrome Decomposition/Solution.py

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return n
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for i in range(n // 2 + 1):
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if text[:i] == text[-i:]:
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return 2 + self.longestDecomposition(text[i: -i])
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return 2 + self.longestDecomposition(text[i:-i])
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return 1

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