LeetCode 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: 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.

分析

letters[i]表示 ‘a’+i结尾的最长子链。len记录的是以p[i]结尾的当时最长子链。如果len<=letters[curr] 则表示这些子链已经重复，不用加到res上。
其实只需要记录子链中以26个字母结尾的长度分别是多少，最后把它们加起来就可以了

class Solution {
public:
    int findSubstringInWraproundString(string p) {
        vector<int> letters(26, 0);
        int res = 0, len = 0;
        for (int i = 0; i < p.size(); i++) {
            int cur = p[i] - 'a';
            if (i > 0 && p[i - 1] != (cur + 26 - 1) % 26 + 'a') len = 0;
            if (++len > letters[cur]) {
                res += len - letters[cur];
                letters[cur] = len;
            }
        }
        return res;
    }
};