Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.
方法1 Dynamic programming
定义一个二维数组记录是否是Palindromic。substring i to j is palindromic==>table[i][j]=1, else table[i][j]=0
base: table[i][i]=1;
table[i][i+1]=1 if s.charAt(i)==s.charAt(i+1);
When substring‘s length>=3:
table[i][j]=1 =>table[i+1][j-1]==1 && s.charAt(i)==s.charAt(j)
Time O(n^2) Space O(n^2)
public class Solution { public String longestPalindrome(String s) { if(s==null||s.length()<=1) return s; int max=0;//max length int len=s.length(); int[][] table=new int[len][len]; String longestPalin=""; for(int i=0;i<len;i++){ table[i][i]=1;//every single char is palindromic if(i==len-1){//length 2 need to stop here break; } if(s.charAt(i)==s.charAt(i+1)){ table[i][i+1]=1; max=2; longestPalin=s.substring(i,i+2); } } for(int l=3;l<=len;l++){ for(int i=0;i<=len-l;i++){ //be careful i<=len-1 e.g. l=3 len=3 int j=i+l-1; if(s.charAt(i)==s.charAt(j)){ table[i][j]=table[i+1][j-1]; if(table[i][j]==1&&l>max){ max=l; longestPalin=s.substring(i,j+1); } }else{ table[i][j]=0; } } } return longestPalin; } }
方法二：
以每个char为中心和以每两个char之间为中心，向两边扩算寻找。
Time O(n^2) Space O(1)
注意最后一行，return s.substring(start+1,end)
input s,i,i 肯定进入循环，循环停止时，start多减了1，end多加了1，所以（start,end+1）变成(start+1,end)
input s,i,i+1, when s.charAt(i)!=s.charAt(i+1),不进入循环，strat+1=end, return ""。进入循环的同上。
public class Solution { public String longestPalindrome(String s) { if(s==null||s.length()<=1) return s; int max=0;//max length int len=s.length(); String longestPalin=""; for(int i=0;i<len;i++){ //center is i String sublen=longestPalindrome(s,i,i); if(sublen.length()>max){ max=sublen.length(); longestPalin=sublen; } //center is between i and i+1 if(i==len-1) break; sublen=longestPalindrome(s,i,i+1); if(sublen.length()>max){ max=sublen.length(); longestPalin=sublen; } } return longestPalin; } public String longestPalindrome(String s, int start, int end){ while(start>=0&&end<s.length()&&s.charAt(start)==s.charAt(end)){ start--; end++; } return s.substring(start+1,end); } }
Longest Palindromic Substring,布布扣,bubuko.com
原文：http://www.cnblogs.com/qingyezi/p/3816110.html