原题链接在这里：https://leetcode.com/problems/unique-substrings-in-wraparound-string/description/
题目：
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.
题解：
以题中"zab"为例. 以‘z‘为尾的最长连续substring长度是1. 只有一个substring "z".
以‘a‘为尾的最长连续substring长度是2. "za" 和 "a".
以‘b‘为尾的最长连续substring长度是3. "zab", "ab" 和 "b".
最长连续以这个letter 结尾的substring长度恰巧是max number of unique substring ends with这个letter.
如果后面有overlap也没有关系只要维护住最长的长度.
最后每个letter的max number of unique substring的和就是答案.
Time Complexity: O(p.length()). Space: O(1).
AC Java:
1 class Solution {
2 public int findSubstringInWraproundString(String p) {
3 int [] dp = new int[26];
4 int maxCount = 0;
5 for(int i = 0; i<p.length(); i++){
6 if(i>0 && (p.charAt(i)-p.charAt(i-1)==1 || p.charAt(i-1)-p.charAt(i)==25)){
7 maxCount++;
8 }else{
9 maxCount = 1;
10 }
11
12 dp[p.charAt(i)-‘a‘] = Math.max(dp[p.charAt(i)-‘a‘], maxCount);
13 }
14
15 int sum = 0;
16 for(int n : dp){
17 sum += n;
18 }
19 return sum;
20 }
21 }