[LeetCode] 467. Unique Substrings in Wraparound String

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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字符串中能形成的不同非空子串的数量。通过巧妙利用字母间的顺序关系，该算法能在O(n)的时间复杂度内解决问题。

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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.

思路分析

这个思路稍微有点难，首先要搞清楚的是，adcd这个字符串，有四种唯一的以d结尾的子字符串。因此我们只需要考虑的就是以某个字母结尾的子字符串有多少种，然后相加即可。
如果出现更长的子字符串，只需要减去原有的记录的，加上这个差值即可。

时间复杂度分析

时间复杂度只是遍历一遍字符串，O(n)

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
    int findSubstringInWraproundString(string p) {
        vector<int> end_letter(26, 0);
        int length =0;
        int result = 0;
        for (int i = 0; i < p.size(); i++) {
            if (i > 0 && p[i-1] != ((p[i] + 26 -1) % 26 - 'a') ) length = 0;
            if (++length > end_letter[p[i] - 'a']) {
                end_letter[p[i] - 'a'] = length;
                result += length - end_letter[p[i]- 'a'];
            }
        }
        return result;
    }
};