feat: add solutions to lc problem: No.0214

No.0214.Shortest Palindrome

Author: yanglbme

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

@@ -201,43 +201,192 @@ impl Solution {
 #### C#
 
 ```cs
-// https://leetcode.com/problems/shortest-palindrome/
-
-using System.Text;
-
-public partial class Solution
-{
-    public string ShortestPalindrome(string s)
-    {
-        for (var i = s.Length - 1; i >= 0; --i)
-        {
-            var k = i;
-            var j = 0;
-            while (j < k)
-            {
-                if (s[j] == s[k])
-                {
-                    ++j;
-                    --k;
-                }
-                else
-                {
-                    break;
-                }
+public class Solution {
+    public string ShortestPalindrome(string s) {
+        int baseValue = 131;
+        int mul = 1;
+        int mod = (int)1e9 + 7;
+        int prefix = 0, suffix = 0;
+        int idx = 0;
+        int n = s.Length;
+
+        for (int i = 0; i < n; ++i) {
+            int t = s[i] - 'a' + 1;
+            prefix = (int)(((long)prefix * baseValue + t) % mod);
+            suffix = (int)((suffix + (long)t * mul) % mod);
+            mul = (int)(((long)mul * baseValue) % mod);
+            if (prefix == suffix) {
+                idx = i + 1;
             }
-            if (j >= k)
-            {
-                var sb = new StringBuilder(s.Length * 2 - i - 1);
-                for (var l = s.Length - 1; l >= i + 1; --l)
-                {
-                    sb.Append(s[l]);
-                }
-                sb.Append(s);
-                return sb.ToString();
+        }
+
+        if (idx == n) {
+            return s;
+        }
+
+        return new string(s.Substring(idx).Reverse().ToArray()) + s;
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- solution:start -->
+
+### 方法二：KMP 算法
+
+根据题目描述，我们需要将字符串 $s$ 反转，得到字符串 $\textit{rev}$，然后求出字符串 $rev$ 的后缀与字符串 $s$ 的前缀的最长公共部分。我们可以使用 KMP 算法，将字符串 $s$ 与字符串 $rev$ 连接起来，求出其最长前缀与最长后缀的最长公共部分。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def shortestPalindrome(self, s: str) -> str:
+        t = s + "#" + s[::-1] + "$"
+        n = len(t)
+        next = [0] * n
+        next[0] = -1
+        i, j = 2, 0
+        while i < n:
+            if t[i - 1] == t[j]:
+                j += 1
+                next[i] = j
+                i += 1
+            elif j:
+                j = next[j]
+            else:
+                next[i] = 0
+                i += 1
+        return s[::-1][: -next[-1]] + s
+```
+
+#### Java
+
+```java
+class Solution {
+    public String shortestPalindrome(String s) {
+        char[] t = (s + "#" + new StringBuilder(s).reverse().toString() + "$").toCharArray();
+        int n = t.length();
+        int[] next = new int[n];
+        next[0] = -1;
+        for (int i = 2, j = 0; i < n;) {
+            if (t[i - 1] == t[j]) {
+                next[i++] = ++j;
+            } else if (j > 0) {
+                j = next[j];
+            } else {
+                next[i++] = 0;
+            }
+        }
+        return new StringBuilder(s.substring(next[n - 1])).reverse().substring(0, s.length() - next[n - 1]) + s;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    string shortestPalindrome(string s) {
+        string t = s + "#" + string(s.rbegin(), s.rend()) + "$";
+        int n = t.size();
+        int next[n];
+        next[0] = -1;
+        next[1] = 0;
+        for (int i = 2, j = 0; i < n;) {
+            if (t[i - 1] == t[j]) {
+                next[i++] = ++j;
+            } else if (j > 0) {
+                j = next[j];
+            } else {
+                next[i++] = 0;
             }
         }
+        return string(s.rbegin(), s.rbegin() + s.size() - next[n - 1]) + s;
+    }
+};
+```
+
+#### Go
 
-        return string.Empty;
+```go
+func shortestPalindrome(s string) string {
+	t := s + "#" + reverse(s) + "$"
+	n := len(t)
+	next := make([]int, n)
+	next[0] = -1
+	for i, j := 2, 0; i < n; {
+		if t[i-1] == t[j] {
+			j++
+			next[i] = j
+			i++
+		} else if j > 0 {
+			j = next[j]
+		} else {
+			next[i] = 0
+			i++
+		}
+	}
+	return reverse(s)[:len(s)-next[n-1]] + s
+}
+
+func reverse(s string) string {
+	t := []byte(s)
+	for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
+		t[i], t[j] = t[j], t[i]
+	}
+	return string(t)
+}
+```
+
+#### TypeScript
+
+```ts
+function shortestPalindrome(s: string): string {
+    const rev = s.split('').reverse().join('');
+    const t = s + '#' + rev + '$';
+    const n = t.length;
+    const next: number[] = Array(n).fill(0);
+    next[0] = -1;
+    for (let i = 2, j = 0; i < n; ) {
+        if (t[i - 1] === t[j]) {
+            next[i++] = ++j;
+        } else if (j > 0) {
+            j = next[j];
+        } else {
+            next[i++] = 0;
+        }
+    }
+    return rev.slice(0, -next[n - 1]) + s;
+}
+```
+
+#### C#
+
+```cs
+public class Solution {
+    public string ShortestPalindrome(string s) {
+        char[] t = (s + "#" + new string(s.Reverse().ToArray()) + "$").ToCharArray();
+        int n = t.Length;
+        int[] next = new int[n];
+        next[0] = -1;
+        for (int i = 2, j = 0; i < n;) {
+            if (t[i - 1] == t[j]) {
+                next[i++] = ++j;
+            } else if (j > 0) {
+                j = next[j];
+            } else {
+                next[i++] = 0;
+            }
+        }
+        return new string(s.Substring(next[n - 1]).Reverse().ToArray()).Substring(0, s.Length - next[n - 1]) + s;
     }
 }
 ```