leetcode:Unique Substrings in Wraparound String

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", sos will look like this: "...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....".

Now we have another string p. Your job is to find out how many unique non-empty substrings ofp are present in s. In particular, your input is the stringp and you need to output the number of different non-empty substrings ofp 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.

这道题的思路是判断给定字符串中的连续非空子串，方法是将字母看成26进制数，然后再判断每一位与之前的数字是否连续。有一点需要特别注意，即当避免之前求出的子串再次计入结果，方法是利用一个数组来记录每个字母结尾对应的最大子串，每次运算一位后都与该数组对比，若发现小于等于数组的值即说明已经计算过了，就不再计入结果。


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';                                                                       //将字母转化为数字
            if (i > 0 && p[i - 1] != (cur + 26 - 1) % 26 + 'a') len = 0;             //若前一个数字不等于当前数字减一，重置长度
            if (++len > letters[cur]) {                                                            //len + 1后还小于letters[cur]说明当前的子串之前已经计算过了，属于重复，因此不再计入
                res += len - letters[cur];
                letters[cur] = len;
            }
        }
        return res;
    }