解题10（LongestSubStrBetweenAB）

题目描述

查找两个字符串a,b中的最长公共子串。若有多个，输出在较短串中最先出现的那个。

输入描述:

输入两个字符串

输出描述:

返回重复出现的字符

示例1

输入

abcdefghijklmnop
abcsafjklmnopqrstuvw

输出

jklmnop

代码如下:

 package com.yzh.hehe;

 import java.util.Scanner;

 public class LongestSubStrBetweenAB {

     public static void main(String[] args) {
         Scanner scanner=new Scanner(System.in);
         while (scanner.hasNext()) {
             String a=scanner.nextLine();
             String b=scanner.nextLine();
             System.out.println(longestSubBetweenAB2(a, b));
         }
         scanner.close();

     }

     private static String longestSubBetweenAB(String a, String  b) {
         int alength=a.length();
         int blength=b.length();
         //保证a串为较短串
         if (alength>blength) {
             String temp=a;
             int tempLength=alength;
             alength=blength;
             blength=tempLength;
             a=b;
             b=temp;
         }
         //从较短串整串开始，长度依次递减从左向右得到较短串子串，如果较长串中包含此子串，即为目标子串
         for (int  i=alength;  i> 0; i--) {
             int count=alength-i+1;
             for (int j = 0; j < count; j++) {
                 if (b.contains(a.substring(j, j+i))) {
                     return a.substring(j,j+i);
                 }
             }
         }
         return null;
     }

     //    动态规划解决最长子串问题
     private static  String longestSubBetweenAB2(String a,String b) {
         int alength=a.length();
         int blength=b.length();
         //保证a串为较短串
         if (alength>blength) {
             String temp=a;
             int tempLength=alength;
             alength=blength;
             blength=tempLength;
             a=b;
             b=temp;
         }
          int[][]arr=new int[alength+1][blength+1];
          String[][] sArr=new String[alength+1][blength+1];
          StringBuilder stringBuilder=new StringBuilder();
          String resultString=null;
          int result=0;
          for (int i = 0; i <= alength; i++) {
             for (int j = 0; j <= blength; j++) {
                 if (i==0||j==0) {
                     arr[i][j]=0;
                     sArr[i][j]="";
                 }else if (a.charAt(i-1)==b.charAt(j-1)) {
                     arr[i][j]=arr[i-1][j-1]+1;
 //                    stringBuilder.append(b.charAt(j-1));
                     sArr[i][j]=sArr[i-1][j-1]+a.charAt(i-1);
                     if (arr[i][j]>result) {
                         result=arr[i][j];
                         resultString=sArr[i][j];
                     }

                 }else {
                     arr[i][j]=0;
                     sArr[i][j]="";
 //                    stringBuilder.delete(0, stringBuilder.length());
                 }
             }
         }
          if (resultString.length()==0) {
             return null;
         }
          return resultString;

     }
 }