Leetcode #5 longest palindromic Substring (M)

[Problem]

Given a string S, find the longest palindromic substring in s. The maximum length of S is assume, and there exists one unique longest palindromic substring.

[Analysis]

There are many ways of thinking about this problem, and there are various discussions on the Internet. What I'm using here is someone who feels better. A method using dynamic programming is used to solve the problem, and the time and space complexity are O (n^2). DP[I][J] is true when the substring from I to J is recorded as Palindromic,base case is I >= J. Then, for each location, whether the substring from the I-start length k is palindromic.

For S[I...J] There are two things:

1. s[i]! = s[j], then dp[i][j] = false;

2. s[i] = = S[j], then dp[i][j] = dp[i + 1][j-1]

Note that in the process of nested loop solving, K outer loop, that is, each loop is actually to solve the K-length sub-string for each position is palindromic, so as to ensure that the solution Dp[i][j] Dp[i + 1][j-1] has a solution.

Another simple idea is to calculate the length of the longest back text string centered on each possible center point. This center point has N-1 + N = 2N, and the time complexity is also O (n^2). Manacher's algorithm can get the optimal time complexity O (n), but it is difficult to understand.

Reference:

Leetcode (5) Longest palindromic substring:http://blog.csdn.net/feliciafay/article/details/16984031

Longest palindromic Substring part ii:http://articles.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html

[Solution]

 Public classSolution { PublicString Longestpalindrome (String s)//Dynamic Programming Solution    {          if(s.length () = = 0)          {              return""; }          if(s.length () = = 1)          {              returns; }            /** DP denotes if a substring s[i: j] is a palindromic*/        Boolean[] DP =New Boolean[S.length ()][s.length ()]; inti,j; /** Initialize DP*/         for(i = 0; i < s.length (); i++)          {               for(j = 0; J < S.length (); j + +)              {                  if(I >=j) {/** if i = = J, it is a palindromic substring of length 1;                        * If I > J, regard it as an empty substring when initializing. */Dp[i][j]=true; }                  Else                  {                      /** Otherwise, initialize to False*/Dp[i][j]=false; }              }          }                    intK; intMaxLen = 1; intStart = 0, end = 0;  for(k = 1; k < s.length (); k++)          {               /** Test If a substring starts from s[i] of length k was palindromic * Note that we have the B ASE case when k = 1 It 's always palindromic*/             for(i = 0; k + i < s.length (); i++) {J= i +K; if(S.charat (i)! =S.charat (j)) {Dp[i][j]=false; }                  Else{Dp[i][j]= Dp[i+1][j-1]; if(Dp[i][j]) {if(k + 1 >maxlen) {MaxLen= k + 1; Start=i; End=J; }                      }                  }              }          }          returnS.substring (Start, end + 1); }}

Leetcode #5 longest palindromic Substring (M)