467 Unique Substrings in Wraparound String-优快云博客

本文探讨了一种算法，用于计算一个特定的无限循环字符串（由字母'a'到'z'组成）中，给定字符串p的所有唯一非空子串出现的次数。通过分析连续子串的特性，提出了一种高效解决方案，利用数组和哈希表来记录并避免重复计数。

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1 题目

Consider the string s to be the infinite wraparound string of "abcdefghijklmnopqrstuvwxyz", so swill 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.

2 尝试解

2.1 分析

给定由小写字母组成的字符串S，问S的所有子串集合中，不重复的连续的子串有多少个。连续指的是s[i]-s[i-1]==1 || s[i]-s[i-1]==-25。

创建与字符串同样大小的数组Count，Count[i]表示以字符S[i]结尾的连续字串的个数（不去重），则

Count[i] = 1 + ((s[i]-s[i-1]==1 || s[i]-s[i-1]==-25)?Count[i-1]:0)

然后考虑重复的问题，用map<char,int>记录以某个小写字母结尾的最大连续子串的长度，不断更新，最后求和即可。

2.2 代码

class Solution {
public:
    int findSubstringInWraproundString(string p) {
        if(!p.size()) return 0;
        vector<int> count(26,0);
        count[p[0]-'a'] = 1;
        vector<int> saver(p.size(),1);
        for(int i = 1; i < p.size(); i ++){
            if(p[i]-p[i-1]==1 || p[i]-p[i-1] == -25)
                saver[i] += saver[i-1];
            count[p[i]-'a'] = max(count[p[i]-'a'],saver[i]);
        }
        int result = 0;
        for(auto record : count)
            result += record;
        return result;
    }
};

3 标准解

class Solution {
public:
    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';
            if (i > 0 && p[i - 1] != (cur + 26 - 1) % 26 + 'a') len = 0;
            if (++len > letters[cur]) {
                res += len - letters[cur];
                letters[cur] = len;
            }
        }
        return res;
    }
};