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feat: add solutions to lc problem: No.0214
No.0214.Shortest Palindrome
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‎README.md

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@@ -61,6 +61,7 @@
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- [和至少为 K 的最短子数组](/solution/0800-0899/0862.Shortest%20Subarray%20with%20Sum%20at%20Least%20K/README.md) - `单调队列`
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- [带限制的子序列和](/solution/1400-1499/1425.Constrained%20Subsequence%20Sum/README.md) - `动态规划`、`单调队列优化`
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- [单词规律 II](/solution/0200-0299/0291.Word%20Pattern%20II/README.md) - `哈希表`、`回溯`
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- [最短回文串](/solution/0200-0299/0214.Shortest%20Palindrome/README.md) - `字符串哈希`
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### 3. 搜索
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‎README_EN.md

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@@ -59,6 +59,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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- [Shortest Subarray with Sum at Least K](/solution/0800-0899/0862.Shortest%20Subarray%20with%20Sum%20at%20Least%20K/README_EN.md) - `Monotonic Queue`
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- [Constrained Subsequence Sum](/solution/1400-1499/1425.Constrained%20Subsequence%20Sum/README_EN.md) - `Dynamic Programming`, `Monotonic Queue`
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- [Word Pattern II](/solution/0200-0299/0291.Word%20Pattern%20II/README_EN.md) - `Hash Table`、`Backtracking`
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- [Shortest Palindrome](/solution/0200-0299/0214.Shortest%20Palindrome/README_EN.md) - `Rabin-Karp`
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### 3. Search
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‎solution/0200-0299/0214.Shortest Palindrome/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：字符串哈希**
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问题等价于**找到字符串 s 的最长回文前缀**。
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记 s 的长度为 n，其最长回文前缀的长度为 m，将 s 的后 n-m 个字符反序并添加到 s 的前面即可构成最短回文串。
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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 shortestPalindrome(self, s: str) -> str:
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base = 131
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mod = 10**9 + 7
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n = len(s)
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prefix = suffix = 0
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mul = 1
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idx = 0
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for i, c in enumerate(s):
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prefix = (prefix * base + (ord(c) - ord('a') + 1)) % mod
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suffix = (suffix + (ord(c) - ord('a') + 1) * mul) % mod
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mul = (mul * base) % mod
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if prefix == suffix:
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idx = i + 1
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return s if idx == n else s[idx:][::-1] + s
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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 String shortestPalindrome(String s) {
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int base = 131;
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int mul = 1;
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int mod = (int) 1e9 + 7;
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int prefix = 0, suffix = 0;
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int idx = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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int t = s.charAt(i) - 'a' + 1;
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prefix = (int) (((long) prefix * base + t) % mod);
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suffix = (int) ((suffix + (long) t * mul) % mod);
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mul = (int) (((long) mul * base) % mod);
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if (prefix == suffix) {
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idx = i + 1;
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}
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}
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if (idx == n) {
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return s;
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}
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return new StringBuilder(s.substring(idx)).reverse().toString() + s;
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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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typedef unsigned long long ull;
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class Solution {
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public:
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string shortestPalindrome(string s) {
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int base = 131;
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ull mul = 1;
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ull prefix = 0;
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ull suffix = 0;
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int idx = 0, n = s.size();
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for (int i = 0; i < n; ++i)
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{
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int t = s[i] - 'a' + 1;
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prefix = prefix * base + t;
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suffix = suffix + mul * t;
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mul *= base;
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if (prefix == suffix) idx = i + 1;
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}
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if (idx == n) return s;
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string x = s.substr(idx, n - idx);
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reverse(x.begin(), x.end());
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return x + s;
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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 shortestPalindrome(s string) string {
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n := len(s)
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base, mod := 131, int(1e9)+7
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prefix, suffix, mul := 0, 0, 1
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idx := 0
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for i, c := range s {
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t := int(c-'a') + 1
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prefix = (prefix*base + t) % mod
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suffix = (suffix + t*mul) % mod
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mul = (mul * base) % mod
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if prefix == suffix {
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idx = i + 1
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}
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}
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if idx == n {
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return s
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}
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x := []byte(s[idx:])
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for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
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x[i], x[j] = x[j], x[i]
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}
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return string(x) + s
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}
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```
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### **...**

‎solution/0200-0299/0214.Shortest Palindrome/README_EN.md

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### **Python3**
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```python
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class Solution:
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def shortestPalindrome(self, s: str) -> str:
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base = 131
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mod = 10**9 + 7
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n = len(s)
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prefix = suffix = 0
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mul = 1
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idx = 0
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for i, c in enumerate(s):
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prefix = (prefix * base + (ord(c) - ord('a') + 1)) % mod
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suffix = (suffix + (ord(c) - ord('a') + 1) * mul) % mod
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mul = (mul * base) % mod
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if prefix == suffix:
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idx = i + 1
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return s if idx == n else s[idx:][::-1] + s
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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 String shortestPalindrome(String s) {
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int base = 131;
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int mul = 1;
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int mod = (int) 1e9 + 7;
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int prefix = 0, suffix = 0;
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int idx = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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int t = s.charAt(i) - 'a' + 1;
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prefix = (int) (((long) prefix * base + t) % mod);
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suffix = (int) ((suffix + (long) t * mul) % mod);
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mul = (int) (((long) mul * base) % mod);
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if (prefix == suffix) {
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idx = i + 1;
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}
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}
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if (idx == n) {
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return s;
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}
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return new StringBuilder(s.substring(idx)).reverse().toString() + s;
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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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typedef unsigned long long ull;
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class Solution {
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public:
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string shortestPalindrome(string s) {
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int base = 131;
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ull mul = 1;
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ull prefix = 0;
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ull suffix = 0;
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int idx = 0, n = s.size();
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for (int i = 0; i < n; ++i)
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{
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int t = s[i] - 'a' + 1;
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prefix = prefix * base + t;
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suffix = suffix + mul * t;
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mul *= base;
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if (prefix == suffix) idx = i + 1;
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}
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if (idx == n) return s;
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string x = s.substr(idx, n - idx);
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reverse(x.begin(), x.end());
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return x + s;
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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 shortestPalindrome(s string) string {
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n := len(s)
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base, mod := 131, int(1e9)+7
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prefix, suffix, mul := 0, 0, 1
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idx := 0
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for i, c := range s {
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t := int(c-'a') + 1
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prefix = (prefix*base + t) % mod
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suffix = (suffix + t*mul) % mod
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mul = (mul * base) % mod
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if prefix == suffix {
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idx = i + 1
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}
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}
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if idx == n {
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return s
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}
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x := []byte(s[idx:])
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for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
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x[i], x[j] = x[j], x[i]
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}
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return string(x) + s
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}
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```
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### **...**

‎solution/0200-0299/0214.Shortest Palindrome/Solution.cpp

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typedef unsigned long long ull;
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class Solution {
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public:
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string shortestPalindrome(string s) {
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int base = 131;
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ull mul = 1;
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ull prefix = 0;
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ull suffix = 0;
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int idx = 0, n = s.size();
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for (int i = 0; i < n; ++i)
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{
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int t = s[i] - 'a' + 1;
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prefix = prefix * base + t;
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suffix = suffix + mul * t;
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mul *= base;
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if (prefix == suffix) idx = i + 1;
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}
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if (idx == n) return s;
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string x = s.substr(idx, n - idx);
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reverse(x.begin(), x.end());
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return x + s;
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}
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};

‎solution/0200-0299/0214.Shortest Palindrome/Solution.go

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func shortestPalindrome(s string) string {
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n := len(s)
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base, mod := 131, int(1e9)+7
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prefix, suffix, mul := 0, 0, 1
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idx := 0
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for i, c := range s {
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t := int(c-'a') + 1
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prefix = (prefix*base + t) % mod
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suffix = (suffix + t*mul) % mod
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mul = (mul * base) % mod
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if prefix == suffix {
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idx = i + 1
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}
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}
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if idx == n {
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return s
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}
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x := []byte(s[idx:])
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for i, j := 0, len(x)-1; i < j; i, j = i+1, j-1 {
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x[i], x[j] = x[j], x[i]
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}
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return string(x) + s
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}

‎solution/0200-0299/0214.Shortest Palindrome/Solution.java

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class Solution {
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public String shortestPalindrome(String s) {
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int base = 131;
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int mul = 1;
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int mod = (int) 1e9 + 7;
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int prefix = 0, suffix = 0;
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int idx = 0;
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int n = s.length();
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for (int i = 0; i < n; ++i) {
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int t = s.charAt(i) - 'a' + 1;
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prefix = (int) (((long) prefix * base + t) % mod);
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suffix = (int) ((suffix + (long) t * mul) % mod);
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mul = (int) (((long) mul * base) % mod);
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if (prefix == suffix) {
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idx = i + 1;
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}
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}
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if (idx == n) {
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return s;
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}
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return new StringBuilder(s.substring(idx)).reverse().toString() + s;
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}
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}

‎solution/0200-0299/0214.Shortest Palindrome/Solution.py

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class Solution:
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def shortestPalindrome(self, s: str) -> str:
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base = 131
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mod = 10**9 + 7
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n = len(s)
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prefix = suffix = 0
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mul = 1
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idx = 0
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for i, c in enumerate(s):
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prefix = (prefix * base + (ord(c) - ord('a') + 1)) % mod
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suffix = (suffix + (ord(c) - ord('a') + 1) * mul) % mod
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mul = (mul * base) % mod
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if prefix == suffix:
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idx = i + 1
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return s if idx == n else s[idx:][::-1] + s

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