467. 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: 1Explanation: Only the substring "a" of string "a" is in the string s.

Example 2:

Input: "cac"Output: 2Explanation: There are two substrings "a", "c" of string "cac" in the string s.

Example 3:

Input: "zab"Output: 6Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

package l467;/* * 扫描一遍，计算以每个字母开头最长能持续多长 */public class Solution {    public int findSubstringInWraproundString(String p) {        char[] cs = p.toCharArray();        int[] startFrom = new int[26];        for(int i=0; i<cs.length; i++) {        int j = i;        while(j+1 < cs.length && ((cs[j]=='z'&&cs[j+1]=='a') || (cs[j]+1==cs[j+1])))         j++;                while(i<=j){        startFrom[cs[i]-'a'] = Math.max(j-i+1, startFrom[cs[i]-'a']);        i++;        }        i--;        }                int ret = 0;        for(int j : startFrom)        ret += j;                return ret;    }}