@@ -201,43 +201,193 @@ impl Solution {
201201#### C#
202202
203203``` cs
204- // https://leetcode.com/problems/shortest-palindrome/
205-
206- using System .Text ;
207- 208- public partial class Solution
209- {
210- public string ShortestPalindrome (string s )
211- {
212- for (var i = s .Length - 1 ; i >= 0 ; -- i )
213- {
214- var k = i ;
215- var j = 0 ;
216- while (j < k )
217- {
218- if (s [j ] == s [k ])
219- {
220- ++ j ;
221- -- k ;
222- }
223- else
224- {
225- break ;
226- }
204+ public class Solution {
205+ public string ShortestPalindrome (string s ) {
206+ int baseValue = 131 ;
207+ int mul = 1 ;
208+ int mod = (int )1 e 9 + 7 ;
209+ int prefix = 0 , suffix = 0 ;
210+ int idx = 0 ;
211+ int n = s .Length ;
212+ 213+ for (int i = 0 ; i < n ; ++ i ) {
214+ int t = s [i ] - 'a' + 1 ;
215+ prefix = (int )(((long )prefix * baseValue + t ) % mod );
216+ suffix = (int )((suffix + (long )t * mul ) % mod );
217+ mul = (int )(((long )mul * baseValue ) % mod );
218+ if (prefix == suffix ) {
219+ idx = i + 1 ;
227220 }
228- if (j >= k )
229- {
230- var sb = new StringBuilder (s .Length * 2 - i - 1 );
231- for (var l = s .Length - 1 ; l >= i + 1 ; -- l )
232- {
233- sb .Append (s [l ]);
234- }
235- sb .Append (s );
236- return sb .ToString ();
221+ }
222+ 223+ if (idx == n ) {
224+ return s ;
225+ }
226+ 227+ return new string (s .Substring (idx ).Reverse ().ToArray ()) + s ;
228+ }
229+ }
230+ ```
231+ 232+ <!-- tabs: end -->
233+ 234+ <!-- solution: end -->
235+ 236+ <!-- solution: start -->
237+ 238+ ### 方法二:KMP 算法
239+ 240+ 根据题目描述,我们需要将字符串 $s$ 反转,得到字符串 $\textit{rev},ドル然后求出字符串 $rev$ 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。
241+ 242+ 时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
243+ 244+ <!-- tabs: start -->
245+ 246+ #### Python3
247+ 248+ ``` python
249+ class Solution :
250+ def shortestPalindrome (self s : str ) -> str :
251+ t = s + " #" + s[::- 1 ] + " $"
252+ n = len (t)
253+ next = [0 ] * n
254+ next [0 ] = - 1
255+ i, j = 2 , 0
256+ while i < n:
257+ if t[i - 1 ] == t[j]:
258+ j += 1
259+ next [i] = j
260+ i += 1
261+ elif j:
262+ j = next [j]
263+ else :
264+ next [i] = 0
265+ i += 1
266+ return s[::- 1 ][: - next [- 1 ]] + s
267+ ```
268+ 269+ #### Java
270+ 271+ ``` java
272+ class Solution {
273+ public String shortestPalindrome (String s ) {
274+ String rev = new StringBuilder (s). reverse(). toString();
275+ char [] t = (s + " #" + rev + " $" . toCharArray();
276+ int n = t. length;
277+ int [] next = new int [n];
278+ next[0 ] = - 1 ;
279+ for (int i = 2 , j = 0 ; i < n;) {
280+ if (t[i - 1 ] == t[j]) {
281+ next[i++ ] = ++ j;
282+ } else if (j > 0 ) {
283+ j = next[j];
284+ } else {
285+ next[i++ ] = 0 ;
286+ }
287+ }
288+ return rev. substring(0 , s. length() - next[n - 1 ]) + s;
289+ }
290+ }
291+ ```
292+ 293+ #### C++
294+ 295+ ``` cpp
296+ class Solution {
297+ public:
298+ string shortestPalindrome(string s) {
299+ string t = s + "#" + string(s.rbegin(), s.rend()) + "$";
300+ int n = t.size();
301+ int next[ n] ;
302+ next[ 0] = -1;
303+ next[ 1] = 0;
304+ for (int i = 2, j = 0; i < n;) {
305+ if (t[ i - 1] == t[ j] ) {
306+ next[ i++] = ++j;
307+ } else if (j > 0) {
308+ j = next[ j] ;
309+ } else {
310+ next[ i++] = 0;
237311 }
238312 }
313+ return string(s.rbegin(), s.rbegin() + s.size() - next[ n - 1] ) + s;
314+ }
315+ };
316+ ```
317+
318+ #### Go
239319
240- return string .Empty ;
320+ ```go
321+ func shortestPalindrome(s string) string {
322+ t := s + "#" + reverse(s) + "$"
323+ n := len(t)
324+ next := make([]int, n)
325+ next[0] = -1
326+ for i, j := 2, 0; i < n; {
327+ if t[i-1] == t[j] {
328+ j++
329+ next[i] = j
330+ i++
331+ } else if j > 0 {
332+ j = next[j]
333+ } else {
334+ next[i] = 0
335+ i++
336+ }
337+ }
338+ return reverse(s)[:len(s)-next[n-1]] + s
339+ }
340+
341+ func reverse(s string) string {
342+ t := []byte(s)
343+ for i, j := 0, len(t)-1; i < j; i, j = i+1, j-1 {
344+ t[i], t[j] = t[j], t[i]
345+ }
346+ return string(t)
347+ }
348+ ```
349+ 350+ #### TypeScript
351+ 352+ ``` ts
353+ function shortestPalindrome(s : string ): string {
354+ const = s .split (' ' reverse ().join (' '
355+ const = s + ' #' + rev + ' $'
356+ const = t .length ;
357+ const : number [] = Array (n ).fill (0 );
358+ next [0 ] = - 1 ;
359+ for (let i = 2 , j = 0 ; i < n ; ) {
360+ if (t [i - 1 ] === t [j ]) {
361+ next [i ++ ] = ++ j ;
362+ } else if (j > 0 ) {
363+ j = next [j ];
364+ } else {
365+ next [i ++ ] = 0 ;
366+ }
367+ }
368+ return rev .slice (0 , - next [n - 1 ]) + s ;
369+ }
370+ ```
371+ 372+ #### C#
373+ 374+ ``` cs
375+ public class Solution {
376+ public string ShortestPalindrome (string s ) {
377+ char [] t = (s + " #" + new string (s .Reverse ().ToArray ()) + " $" ToCharArray ();
378+ int n = t .Length ;
379+ int [] next = new int [n ];
380+ next [0 ] = - 1 ;
381+ for (int i = 2 , j = 0 ; i < n ;) {
382+ if (t [i - 1 ] == t [j ]) {
383+ next [i ++ ] = ++ j ;
384+ } else if (j > 0 ) {
385+ j = next [j ];
386+ } else {
387+ next [i ++ ] = 0 ;
388+ }
389+ }
390+ return new string (s .Substring (next [n - 1 ]).Reverse ().ToArray ()).Substring (0 , s .Length - next [n - 1 ]) + s ;
241391 }
242392}
243393```
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