leetcode-5-Longest Palindromic Substring

题目：Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"



思路：寻找字符串的最长回文串的题，属于老生常谈的题了，一般要么求出长度为多少，要么求出子串，方法都是一样的：动态规划；中心扩散法；马拉车算法；一般就这三种 
算法思想可看我的另一篇博客：最长回文子串
法1：动态规划（时间复杂度O(n^2）,空间复杂度O(n^2))

 1 class Solution {
 2    public static String  longestPalindrome(String s){
　　　　　　if(s==null) return null;
 3        int n=s.length();
 4         int[][]dp=new int[n+1][n+1];
 5         int max=1;
 6         int x=0,y=0;
 7         for(int i=0;i<=n;i++){
 8             dp[i][i]=1;
 9         }
10         for(int len=2;len<=n;len++){
11             //len表示子串长度，i表示子串开头位置，j表示子串结束位置
12             for(int i=0;i<=n-len;i++){
13                 int j=i+len-1;
14                 //开始比较
15                 if(len==2&&s.charAt(i)==s.charAt(j)){
16                     dp[i][j]=2;
17                     max=2;
18                     x=i;
19                     y=j;
20                 }else{
21                     if(s.charAt(i)==s.charAt(j)&&dp[i+1][j-1]!=0){
22                         dp[i][j]=len;
23                         max=len;
24                         x=i;
25                         y=j;
26                     }
27                 }
28             }
29 
30         }
31        return s.substring(x,y+1);
32    }
33 }

这种方法看来并不算太好，Your runtime beats 21.57 % of java submissions.

再用中心扩散法试试
Your runtime beats 93.63 % of java submissions

 1 public class Solution {
 2     int left=0;
 3     int max=1;
 4     public String longestPalindrome(String s) {
 5         char[] cs=s.toCharArray();
 6         for(int i=0;i<cs.length;i++){
 7             findPailndromic(cs,i,i);
 8             findPailndromic(cs,i,i+1);
 9         }
10         return s.substring(left,left+max);
11     }
12     public void findPailndromic(char[]cs,int i,int j){
13         while(i>=0&&j<cs.length&&cs[i]==cs[j]){
14             if(max<j-i+1){
15                 left=i;
16                 max=j-i+1;
17             }
18             i--;
19             j++;
20         }
21     }
22 }