leetcode-5

/*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”
*/

/**
*

• @author administered
• 使用dp进行求解
• path[i][j] 表示从i到j是否为回文串，如果是则为1 ，否则0
• i=j

*/

public class Solution5 {

	 public static String longestPalindrome(String s) {
		 if (s == null || s.length() < 2) {
	            return s;
	        }
		 char[] chs=s.toCharArray();
		 int bp[][]=new int[chs.length][chs.length];
		 int start=0;
		 int end=0;
		 int big=0;
         //初始化
		 for(int i=0;i<chs.length;i++) {
			 bp[i][i]=1;
			 
		  }
		 for(int i=0;i<chs.length-1;i++) {
			 if(chs[i]==chs[i+1]) {
				 bp[i][i+1]=1; 
			 }
			 
		  }
		 
			 for(int l=1;l<chs.length;l++) {
				 for(int i=0;i+l<chs.length;i++) {
					 int j=i+l;
					 System.out.println("i:"+i);
					 System.out.println("j:"+j);
					 System.out.println("chsi:"+chs[i]);
					 System.out.println("chsj:"+chs[j]);
					 System.out.println("bp[i+1][j-1]:"+bp[i+1][j-1]);
					 
					 if(chs[i]==chs[j]&&bp[i+1][j-1]==1) {
						bp[i][j]=1;
						 
					 }
				 }
			 }
			 
			 
		
		 
		 //遍历
		 for(int i=0;i<chs.length;i++) {
			 for(int j=i;j<chs.length;j++) {
				  if(j-i+1>big&&bp[i][j]==1) {
					 big=j-i+1;
					 start=i;
					 end=j;
					
					
				 }
			 } 
		 }
		 
		
		 for(int i=0;i<chs.length;i++) {
			 for(int j=0;j<chs.length;j++) {
				 System.out.print(bp[i][j]+"\t");
				 
			 }
			 System.out.println("");
		 }
		 
		 String r=s.substring(start,end+1);
		 return r;
	
		 
		 
	 }
	 public static void  main(String args[]) {
		 String ch="ccc";
		 String r=Solution5.longestPalindrome(ch);
		 System.out.print(r);
		 
		 
	 }
	 
	 
}