LeetCode Unique Substrings in Wraparound String

Posted 2020-10-08 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/unique-substrings-in-wraparound-string/description/

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.

题解：

以题中"zab"为例. 以‘z‘为尾的最长连续substring长度是1. 只有一个substring "z".

以‘a‘为尾的最长连续substring长度是2. "za" 和 "a".

以‘b‘为尾的最长连续substring长度是3. "zab", "ab" 和 "b".

最长连续以这个letter 结尾的substring长度恰巧是max number of unique substring ends with这个letter.

如果后面有overlap也没有关系只要维护住最长的长度.

最后每个letter的max number of unique substring的和就是答案.

Time Complexity: O(p.length()). Space: O(1).

AC Java:

 1 class Solution {
 2     public int findSubstringInWraproundString(String p) {
 3         int [] dp = new int[26];
 4         int maxCount = 0;
 5         for(int i = 0; i<p.length(); i++){
 6             if(i>0 && (p.charAt(i)-p.charAt(i-1)==1 || p.charAt(i-1)-p.charAt(i)==25)){
 7                 maxCount++;
 8             }else{
 9                 maxCount = 1;
10             }
11             
12             dp[p.charAt(i)-‘a‘] = Math.max(dp[p.charAt(i)-‘a‘], maxCount);
13         }
14         
15         int sum = 0;
16         for(int n : dp){
17             sum += n;
18         }
19         return sum;
20     }
21 }