467. Unique Substrings in Wraparound String

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.

• 这道题目稍微有点难度，关键找出是否存在重复的字串，之前写的用来查找是否存在字串的查找算法，结果超时。

/*解法*/
bool isContinue(char pre,char cur){  
       if(pre == 'z'){
           if(cur == 'a'){
               return true;
           }else{
               return false;
           }
       }else{
           if(cur == pre+1){
               return true;
           }else{
               return false;
           }
       }

       return true;
    }

    int findSubstringInWraproundString(string p) {
        vector<int> letters(26,0);
        int res = 0;
        int len = 0;

        if(p.size() <= 0){
            return 0;
        }

        letters[p[0]-'a'] = 1;
        len = 1;
        res++;

        for(int i = 1; i < p.size();++i){           
            if(!isContinue(p[i-1],p[i])){
                len = 0;         
            }

            len++;
            if(len > letters[p[i]-'a']){
                res += (len - letters[p[i]-'a']);
                letters[p[i]-'a'] = len;               
            }
        }

        return res;
    }

/*自己写的算法超时*/
class Solution {
public:
    bool isContinue(char pre,char cur){  
       if(pre == 'z'){
           if(cur == 'a'){
               return true;
           }else{
               return false;
           }
       }else{
           if(cur == pre+1){
               return true;
           }else{
               return false;
           }
       }

       return true;
    }

    int findSubstringInWraproundString(string p) {
        int n = p.size();
        set<int> s;
        vector<set<int>> cnt(26,s);
        vector<int> len(n,0);
        vector<int> dp(n,0);

        if(n <= 0){
            return 0;
        }

        len[0] = 1;
        dp[0] = 1;
        cnt[p[0]-'a'].insert(1);

        for(int i = 1;i < n;++i){
            dp[i] = dp[i-1];
            if(isContinue(p[i-1],p[i])){
                len[i] = len[i-1] + 1;        
            }else{
                len[i] = 1;
            }

            for(int j = 1;j <= len[i]; ++j){
                int target = p[i-j+1]-'a';
                if(cnt[target].find(j) == cnt[target].end()){
                    cnt[target].insert(j);
                    dp[i]++;
                }//not find the sub string
            }       
        }

        return dp[n-1];
    }
};