[Leetcode] Longest palindromic Substring

This problem have a long story. There is just too many solutions on the web and it is can studied for a long time before you fully grasp it. Morever, it can induce many other concepts or problems (longest palindromic subsequence, longest common substring, etc).

The simplest-solve it is-to-use two-dimensional DP. We denote p[i][j] to being an indicator of whether the substring from i to J (inclusive) is a Pali Ndrome. It is obvious that the following relationships hold:

1. P[i][i] = 1 (Each character itself is palindromic);
2. p[i][i + 1] = s[i] = = S[j] (Neighboring characters is palindromic if they is the same);
3. P[i][j] = p[i + 1][j-1] && s[i] = = S[j] (If the substring is palindrome, then adding the same character at both of it, ends still gives a palindrome).

1 and 2 are base cases and 3 is the general case.

Then we'll have the following unoptimiezd DP code.

1     stringLongestpalindrome (strings) {2         intStart =0, Len =1, n =s.length ();3         BOOLdp[ +][ +] = {false};4          for(inti =0; I < n; i++)5Dp[i][i] =true;6          for(inti =0; I < n-1; i++) {7Dp[i][i +1] = s[i] = = S[i +1];8             if(Dp[i][i +1]) {9Start =i;TenLen =2; One             } A         } -          for(intL =3; L <= N; l++) { -              for(inti =0; I < N-l +1; i++) { the                 intj = i + L-1; -DP[I][J] = dp[i +1][j-1] && s[i] = =S[j]; -                 if(Dp[i][j]) { -Start =i; +Len =l; -                 } +             } A         } at         returns.substr (start, Len); -}

Note that each time if we update dp[i][j], we only need Dp[i + 1][j-1] from the left column, so we C An maintain a, variable for it, and reduce the space complexity from O (n^2) to O (n). The code now becomes as follows.

1     stringLongestpalindrome (strings) {2         intStart =0, Len =1, n =s.length ();3         BOOLcur[ +] = {false};4         BOOLPre;5cur[0] =true;6          for(intj =1; J < N; J + +) {7CUR[J] =true;8Pre = Cur[j-1];9Cur[j-1] = s[j-1] ==S[j];Ten             if(Cur[j-1] && Len <2) { OneStart = J-1; ALen =2; -             } -              for(inti = J-2; I >=0; i--) { the                 BOOLtemp =Cur[i]; -Cur[i] = Pre && s[i] = =S[j]; -                 if(Cur[i] && j-i +1>Len) { -Start =i; +Len = j-i +1; -                 } +Pre =temp; A             } at         } -         returns.substr (start, Len); -}

We may also traverse the string and expand to left and right from any character to obtain the longest palindrome. The following code should is self-explanatory.

1     stringSearchstringSintLeftintRight ) {2         intL = left, r =Right ;3          while(L >=0&& R < s.length () && s[l] = =S[r]) {4l--;5r++;6         }7         returnS.SUBSTR (L +1, R-l-1);8     }9     Ten     stringLongestpalindrome (strings) { One         stringLongest = S.substr (0,1); A          for(inti =0; I < s.length ()-1; i++) { -             stringTMP1 =search (s, I, I); -             stringTMP2 = Search (s, I, i +1); the             if(Tmp1.length () > Longest.length ()) longest =Tmp1; -             if(Tmp2.length () > Longest.length ()) longest =TMP2; -         } -         returnlongest; +}

Of course, this problem still have a non-trivial O (n) algorithm, named Manacher ' s algorithm. This page is a nice explanation for it. The final code is shown below.

1     stringProcessstrings) {2         intn =s.length ();3         stringT2* n +3,'#');4t[0] ='$';5t[2* n +2] ='%';6          for(inti =0; I < n; i++)7t[2* (i +1)] =S[i];8         returnT;9     }Ten      One     stringLongestpalindrome (strings) { A         stringt =process (s); -         intn =t.length (); -         int* Plen =New int[n] (); the         intCenter =0, right =0; -          for(inti =1; I < n-1; i++) { -             intI_mirror =2* Center-i; -Plen[i] = right > I? Min (Plen[i_mirror], right-i):0; +              while(T[i + plen[i] +1] = = T[i-plen[i]-1]) -plen[i]++; +             if(i + plen[i] >Right ) { ACenter =i; atright = i +Plen[i]; -             } -         } -         intMaxLen =0; -          for(inti =1; I < n-1; i++) { -             if(Plen[i] >maxlen) { inCenter =i; -MaxLen =Plen[i]; to             } +         } -         Delete[] Plen; the         returnS.SUBSTR (center-1-MaxLen)/2, maxlen); *}

[Leetcode] Longest palindromic Substring