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;
        
    }
}