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.

Reference

public class Solution {    public int findSubstringInWraproundString(String p) {    // count[i] is the maximum unique substring end with ith letter.        // 0 - 'a', 1 - 'b', ..., 25 - 'z'.        int[] count = new int[26];        int maxLengthCur = 0;                for (int i = 0; i < p.length(); i++) {            int len = 1;            if (i > 0 && (p.charAt(i) - p.charAt(i - 1) == 1 || (p.charAt(i - 1) - p.charAt(i) == 25)))                maxLengthCur++;            else                maxLengthCur = 1;            int index = p.charAt(i) - 'a';            count[index] = Math.max(count[index], maxLengthCur);        }                // Sum to get result        int sum = 0;        for (int i = 0; i < 26; i++) {            sum += count[i];        }        return sum;    }}

总结：这道题没有想出来，想到了最长公共子序列，但是情境不同。没有找到子问题。

count[index] = Math.max(count[index], maxLengthCur);

保证了以某个字母结尾的子序列数最大。

子序列数实际上和长度有关，多少个长度。