LeetCode 467. Unique Substrings in Wraparound String
阿新 • • 發佈:2018-12-04
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結尾的最長子鏈。len記錄的是以p[i]結尾的當時最長子鏈。如果len<=letters[curr] 則表示這些子鏈已經重複,不用加到res上。
其實只需要記錄子鏈中以26個字母結尾的長度分別是多少,最後把它們加起來就可以了
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; } };