[Leetcode] 467. Unique Substrings in Wraparound String 解题报告
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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: 1Explanation: Only the substring "a" of string "a" is in the string s.
Example 2:
Input: "cac"Output: 2Explanation: There are two substrings "a", "c" of string "cac" in the string s.
Example 3:
Input: "zab"Output: 6Explanation: 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; }};
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