467. Unique Substrings in Wraparound String

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本文介绍LeetCode 467题“Unique Substrings in Wraparound String”的解题思路及实现方法，通过动态规划算法高效求解给定字符串在无限循环字母表中的独特子串数量。

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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.

Eample 3:

Input: "zab"
Output: 6
Explanation: 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;
  }
}