[LeetCode] 467. Unique Substrings in Wraparound String

題目鏈接: https://leetcode.com/problems/unique-substrings-in-wraparound-string/description/

Description

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.

解題思路

整體思路爲動態規劃，先舉幾個例子來看看。

從最簡單的例子開始，p = "acb"。p　中沒有長度超過 2 的子串是 s 的子串，只有單字符子串 "a", "c", "b"，共 3 個。

稍微複雜一點點的，p = "abc"。p 正好是 s 的子串，則 p 的任意子串都是 s 的子串，分別爲 "a", "b", "ab", "c", "bc", "abc"，可以看到 a 結尾的子串有 1 個，b 結尾的子串有 2 個，c 結尾的子串有 3 個。

再稍微複雜一點點的，p = "ababc"。此時，p 不是 s 的子串，但是 p 中有一個子串 ab　是另一個子串 abc 的子串，存在包含關係，那麼 p 中能匹配到 s 的非空子串數量由更長的子串 abc 決定。

總結上面這些例子，若 p 中有一個能匹配到 s 的長度爲 t 的連續子串，則以子串中每個字符爲結尾的子串個數分別爲 1, 2, ..., t，並且若有另一個包含該子串的更長的子串，則以更長的子串爲主。

因此，算法的具體實現爲先創建一個長度爲 26 的數組用來記錄每個字符爲結尾的子串最多有多少個，最後將該數組每一項相加即爲結果。

時間複雜度 O(n), 空間複雜度 O(1)。

Code

class Solution {
public:
    int findSubstringInWraproundString(string p) {
        vector<int> alpha(26, 0);
        int maxLen, res = 0;
        for (int i = 0; i < p.size(); ++i) {
            if (i > 0 && (p[i] - p[i - 1] == 1 || p[i - 1] - p[i] == 25))
                maxLen++;
            else
                maxLen = 1;
            alpha[p[i] - 'a'] = max(alpha[p[i] - 'a'], maxLen);
        }

        for (auto x: alpha) res += x;

        return res;
    }
};