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.
有一种很巧妙地想法，因为所求的子串是有序的 ，所以没必要用hash记录每一个出现过的子串 ，只需要记录出现过的有序子串的最末尾字母以及该有序子串的长度即可，所以用一个int[26]就可以满足条件，而便利到最后，数组中所有元素加就是所求：
public class Solution2 { //很巧妙地想法 因为是有序的 所以没必要用hash记录每一个出现过的子串 只需要记录出现过的有序子串的最末尾字母以及该有序子串的长度即可所以用一个int[26]即可 public int findSubstringInWraproundString(String p) { if(p.length()<=1) return p.length(); int[] map=new int[26]; int count=1; map[p.charAt(0)-'a']=1; for(int i=1;i<=p.length();i++){ if(i==p.length()||!isNext(p.charAt(i-1),p.charAt(i))){ count=0; } count++; map[p.charAt(i-1)-'a']=Math.max(map[p.charAt(i-1)-'a'], count); } int re=0; for(int i=0;i<map.length;i++){ re+=map[i]; } return re; } public boolean isNext(char a,char b){ return b==a+1||(a=='z'&&b=='a'); } }