[Leetcode] 467. Unique Substrings in Wraparound String 解题报告

题目：

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：

刚开始的时候竟然没有看懂题目，后来才明白这道题目的本质就是寻找p中“循环连续”的子串数目之和，我这里说的“循环连续”指的就是z和a之间也被视作是连续的，就像a和b之间一样。那么如何做呢？一个很不错的方法就是对于a-z中的每个字符，我们寻找以该字母为结尾（或者为首）的最长循环连续子串。我们最终返回的就是以各个字符为结尾（或者为首）的最长循环连续子串的长度的和。为什么会是循环连续的子串的长度呢？例如对于zabcde，其长度为6，它就对应了e，de，cde，bcde，abcde，zabcde这六个子串），并且该最长长度也囊括了以e结尾的更短一些的子串的情况，例如bcde。

代码：

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') { // now it begins
len = 0;
}
if (++len > letters[cur]) { // now it breaks
res += (len - letters[cur]);
letters[cur] = len;
}
}
return res;
}
};