467. Unique Substrings in Wraparound String

https://leetcode.com/problems/unique-substrings-in-wraparound-string/#/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.

dp.

用一个length数组记录每个字母作为最大子串的最后一个字母时的子串长度。

class Solution {
public:
    int findSubstringInWraproundString(string p) {
        vector<int> length(26,0);
        const int size=p.length();
        int len=0,num=0;
        for(int i=0;i<size;i++)
        {
            int cur=p[i]-'a';
            //比较现在的字母和前一个字母，(x+25)%26可以得到前一个字母（a为0...z为26）
            if(i!=0&&p[i-1]!=(cur+25)%26+'a')
                len=0;
            if(++len>length[cur])//++表示每个字母至少为长度为1的子串的最后一个字母
            {
                num+=len-length[cur];//由于s串固定，所以子串从最后一个字母往前推都是固定的，只是可以推多长不确定
                length[cur]=len;//当前字母作为最后一个字母的子串还可以增长
            }
        }
        return num;
    }
};