Leetcode: Unique Substrings in Wraparound String

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so s will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".

Now we have another string p. Your job is to find out how many unique non-empty substrings of p are present in s. In particular, your input is the string p and you need to output the number of different non-empty substrings of p in the string s.

Note: p consists of only lowercase English letters and the size of p might be over 10000.

Example 1:
Input: "a"
Output: 1

Explanation: Only the substring "a" of string "a" is in the string s.
Example 2:
Input: "cac"
Output: 2
Explanation: There are two substrings "a", "c" of string "cac" in the string s.
Example 3:
Input: "zab"
Output: 6
Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

这道题我一开始发现了数学规律：

length of consecutive chars   vs   number of different substring

“a” -> 1 = 1

"ab" -> 3 = 1+2

"abc" -> 6 = 1+2+3

"abcd" -> 10 = 1+2+3+4

于是打算计算string p里面每一段consecutive substring的length是多少，比如abcxyz, 两段分别为“abc”“xyz”， len分别为3和3，#of substring分别是6和6，总数是6+6。 但是这个方法没法处理有重复的，比如abczab，"abc"依然能产生6个substring, 然而"zab"就有重复了，只能产生3个different substring.

如何处理重复呢？ “abc”和“zab”里面的"ab"要同时考虑到，于是就有了DP的想法：

total number of different substring = sum of {number of different substring end by each char}

Further,

number of different substring end by each char =  length of longest contiguous substring ends with that letter. 

Example "abcd", the  number of unique substring ends with 'd' is 4, apparently they are "abcd", "bcd", "cd" and "d".

If there are overlapping, we only need to consider the longest one because it covers all the possible substrings. Example:"abcdbcd", the max number of unique substring ends with 'd' is 4 and all substrings formed by the 2nd "bcd" part are covered in the 4 substrings already.

 public class Solution {
     public int findSubstringInWraproundString(String p) {
         if (p==null || p.length()==0) return 0;
         int[] nums = new int[26];  //numbers of different substrings end with a specific char

         int longestConsec = 0;
         for (int i=0; i<p.length(); i++) {
             if (i>0 && (p.charAt(i-1)+1 == p.charAt(i) || p.charAt(i-1)-25 == p.charAt(i))) {
                 longestConsec++;
             }
             else longestConsec = 1;

             char c = p.charAt(i);
             nums[(int)(c-'a')] = Math.max(nums[(int)(c-'a')], longestConsec);
         }

         int res = 0;
         for (int num : nums) {
             res += num;
         }
         return res;
     }
 }