LeetCode #467 - Unique Substrings in Wraparound String_leetcode 467 子字符串-优快云博客

本文链接：https://blog.youkuaiyun.com/LawFile/article/details/81256978

本文介绍了一种高效算法，用于计算一个给定字符串在无限循环的英文字母表中，有多少不同的非空子串可以找到。通过分析字符间的连续性，该算法能够快速得出结果，特别适用于处理长度超过10,000的大字符串。

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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:

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.

找出给定字符串的子字符串，满足它们是"...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd...."的子字符串，解决这道题的方法是分别确定以各个字母为结尾的最长子字符串，例如以‘a’为结尾的子字符串的最大程度为k，那么一共有k个以‘a’为结尾的子字符串满足题意。

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