题目描述:

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:

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Input: "a"
Output: 1

Explanation: Only the substring "a" of string "a" is in the string s.

Example 2:

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Input: "cac"
Output: 2
Explanation: There are two substrings "a", "c" of string "cac" in the string s.

Example 3:

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Input: "zab"
Output: 6
Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

动态规划题目. dp[i]记录以p[i]为结尾的符合要求的子串的长度. 为了防止重复, 用另一个数组tail以字母为索引保存结果中以某个字母结尾的最长的子串的长度. 当p[i]与p[i-1]是连续的时候, p[i]=p[i-1]; 否则p[i]=1. 再检查tail[p[i]-'a']与dp[i]的大小关系, 若tail[p[i]-'a']>=dp[i], 说明p[i]结尾的所有子串都已经在结果中了; 否则更新tail[p[i]-'a']=dp[i], 增加结果集中以p[i]这个字母结尾的子串的数量. 最后再对tail数组求和就得到结果. 时间复杂度O(n).

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class Solution {
public:
    int findSubstringInWraproundString(string p) {
        if(p.empty()) return 0;
        vector<int> dp(p.length(), 0);
        vector<int> tail(26, 0);
        dp[0] = 1;
        tail[p[0] - 'a'] = 1;
        for(int i = 1; i < p.length(); i++){
            if((p[i - 1] != 'z' && p[i] == p[i - 1] + 1) || (p[i - 1] == 'z' && p[i] == 'a')){
                dp[i] = dp[i - 1] + 1;
            }
            else{
                dp[i] = 1;
            }
            int index = p[i] - 'a';
            tail[index] = max(tail[index], dp[i]);
        }
        int ans = 0;
        for(auto i : tail){
            ans += i;
        }
        return ans;
    }
};