[LeetCode] 467. Unique Substrings in Wraparound String

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)

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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;
    }
};