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。

 1 class Solution {
 2 public:
 3     int findSubstringInWraproundString(string p) {
 4         vector<int> length(26,0);
 5         const int size=p.length();
 6         int len=0,num=0;
 7         for(int i=0;i<size;i++)
 8         {
 9             int cur=p[i]-‘a‘;
10             //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）
11             if(i!=0&&p[i-1]!=(cur+25)%26+‘a‘)
12                 len=0;
13             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
14             {
15                 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
16                 length[cur]=len;//当前字母作为最后一个字母的子串还可以增长
17             }
18         }
19         return num;
20     }
21 };