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. 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; } }