【KMP】hdu1867（A + B for you again） 杭电java a题真坑

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Problem Description

Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, the result of the addition between them is “asdfg”, for “sdf” is the tail substring of “asdf” and the head substring of the “sdfg” . However, the result comes as “asdfghjk”, when you have to add “asdf” and “ghjk” and guarantee the shortest string first, then the minimum lexicographic second, the same rules for other additions.

 

Input

For each case, there are two strings (the chars selected just form ‘a’ to ‘z’) for you, and each length of theirs won’t exceed 10^5 and won’t be empty.

 

Output

Print the ultimate string by the book.

 

Sample Input

   
   
   
   

    
    
    
    asdf sdfg
asdf ghjk
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    asdfg
asdfghjk
   
   
   
   

解题思路：题目就是字符串模式匹配，注意一些细节问题就可以了。但是java写的就是无限超内存，今天心情不好，再交，就过了，这是无语了啊

import java.util.*;
class P1867{
    static int[] next=new int[100005];
    public static void main(String args[]){
        int n,m,i,len1,len2;
        String str1,str2;
        Scanner sc=new Scanner(System.in);
        while(sc.hasNext()){
            str1=sc.next();
            str2=sc.next();
            len1=str1.length();
            len2=str2.length();
            n=kmp(str1,str2);
            m=kmp(str2,str1);
            if(n==m){
                if(str1.compareTo(str2)>0){
                    System.out.println(str2+str1.substring(n,len1));
                }else{
                    System.out.println(str1+str2.substring(n,len2));
                }
            }else if(n>m){
                System.out.println(str1+str2.substring(n,len2));
            }else{
                System.out.println(str2+str1.substring(m,len1));
            }
        }
    }
    public static void set_next(String str){
        int i=0,j=-1;
        next[0]=-1;
        int len=str.length();
        while(i<len){
            if(j==-1||str.charAt(i)==str.charAt(j)){
                i++;
                j++;
                next[i]=j;///System.out.print(next[i]+" ");
            }else{
                j=next[j];
            }
        }
        //System.out.println();
    }
    public static int kmp(String str1,String str2){
        int i=0,j=0;
        int len1=str1.length(),len2=str2.length();
        set_next(str2);
        while(i<len1){//System.out.print(j+" ");
            if(j==-1||(j<len2&&str1.charAt(i)==str2.charAt(j))){
                i++;
                j++;//System.out.print("** ");
            }else{
                j=next[j];
            }//System.out.print(j+"* ");
        }//System.out.println();
        if(i==len1){
            return j;
        }
        return 0;
    }
}