LeetCode_ 28 (implement strstr())

Title Description ：
Here are two strings haystack and needle , Please come in haystack Find in string needle The first place the string appears （ Subscript from 0 Start ）. If it doesn't exist , Then return to -1 .

explain ：
When needle When it's an empty string , What value should we return ？ This is a good question in an interview .
For this question , When needle When it's an empty string, we should return 0 . This is related to C Linguistic strstr() as well as Java Of indexOf() The definition matches .

Example 1：
Input ：haystack = “hello”, needle = “ll”
Output ：2

Example 2：
Input ：haystack = “aaaaa”, needle = “bba”
Output ：-1

Tips ：
1 <= haystack.length, needle.length <= 104
haystack and needle It only consists of lowercase English characters

class Solution {
    
    public int strStr(String str, String subs) {
    
    	if(str==null||subs==null)return -1;
        if(str.length()==0||subs.length()==0)return -1;
        char[] subc=subs.toCharArray();                 // Convert substrings into character arrays for easy calculation next Array 
        int[] next=new int[subc.length];                // Statement next Array 
        getNext(subc,next);                             // Calculation next Array 
        int i=0,j=0;                                    // Double pointer traverses main string and substring 
        while(i<str.length()&&j<subs.length()){
    
            if(j==-1||str.charAt(i)==subs.charAt(j)){
    
                i++;j++;
            }
            else{
    
                j=next[j];                              // Substring backtracking 
            }
            if(j==subs.length()){
    
                return i-j;                             // return index Subscript 
            }
        }
        return -1;
    }
    public  void getNext(char[] subc,int[] next){
    
        next[0]=-1;                                 // Set the first bit of the array to -1
        int i=1;                                    // Subscript from substring to 2 Start to calculate the position of 
        int j=-1;                                   // here j by next[0] Value 
        while(i<subc.length){
                           // Traversal substring 
            if(j==-1||subc[i-1]==subc[j]){
              // Judge whether the characters after the last longest public prefix are equal 
                next[i++]=++j;                      // Maximum common prefix length plus one 
            }
            else{
    
                j=next[j];                          // Recursively find the longest public prefix forward 
            }
        }
    }
}