467. Unique Substrings in Wraparound String

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: 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.

• Eample 3:

  Input: "zab"Output: 6Explanation: There are six substrings "z", "a", "b", "za", "ab", "zab" of string "zab" in the string s.

• 题目大意：给定一个无限循环的字符串（从A到Z无限循环）S，给定一个字符串P，问P中有多少个非空连续的子串可以在S中出现。

• 思路：DP，首先观察字符串abc，有字符子串a,b,ab,c,bc,abc，其中以a结尾的子串只有a长度为1，以b结尾的子串有b,ab长度为2，以c结尾的子串有c,bc,abc，观察可得，要求以某一字符结尾的子串的个数，即求的以该字符结尾的最长的子串长度即可。所以找出p中每个字符（a-z）结尾的最长连续子串的长度，并将所有长度相加，即所求结果。

• 代码

  package DP;/*** @author OovEver* 2017/12/24 23:53*/public class LeetCode467 {  public int findSubstringInWraproundString(String p) {      int[] count = new int[26];//        以某个字符结尾的字符的最大长度      int maxLength = 0;      for(int i=0;i<p.length();i++) {          if (i > 0 && (p.charAt(i) - p.charAt(i - 1) == 1 || p.charAt(i) - p.charAt(i - 1) == -25)) {              maxLength++;          } else {              maxLength = 1;          }          int index = p.charAt(i) - 'a';          count[index] = Math.max(count[index], maxLength);      }      int sum = 0;      for(int i=0;i<26;i++) {          sum += count[i];      }      return sum;  }}