【LeetCode】467. Unique Substrings in Wraparound String

問題描述

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.

題解

這個題剛看到時，表示不會寫，也不理解題的意思。後來在網上找了題解，研究了半天才弄明白的。現在寫寫自己的見解，希望對自己有所提高。
我們可以發現給定的字符串s，是一個a-z 26個字母的循環。題目要求當輸入一段字符串，要求得輸入的字符串的子串也是給定s的子串的數目；
其實找出以’a-z’每個字符結尾的情況下，最長的子串有多長，然後將其相加就可以；

int findSubstringInWraproundString(string p)
{
    int cnt[26];
    int i = 0;
    for (; i < 26; i++){
        cnt[i] = 0;
    }
    int len = 0;
    int res = 0;
    for (i = 0; i < p.length(); i++){
        int cur = p[i] - 'a';
        if (i > 0 && p[i - 1] != (cur + 26 - 1) % 26 + 'a') len = 0;
        if (++len > cnt[cur]) {
            res += len - cnt[cur];
            cnt[cur] = len;
        }

    }

    return res;
}