Leetcode 467. Unique Substrings in Wraparound String 子串问题 结题报告

1 解题思想

这道题是说给了一个字符串P，需要从其中找到所有出现在S中的子串。
这里的S的子串是一个比较神奇的存在，起就是abcdedf..zabcde…这样无限迭代下去的一个串，也就是只要P的字符串一直递增就好（z->a进位）

所以解题方法也很直接：
找出以’a-z’每个字符结尾的情况下，最长的子串有多长，然后将其相加就可以

2 原题

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.

3 AC解

public class Solution {    /**     * 其实就是找出P中有多少个子串，子串能够在abcdedf...zabcdefg...zab...这样的一个子串里面找到     * 也就是递增的过程     * 考虑为一个26进制的数的序列     * 使用dp的方式，找到以某个字符结尾的最长的有多少种可能     * */    public int findSubstringInWraproundString(String p) {        int p_int[] = new int[p.length()];        int count[] = new int[26];        for(int i=0;i<p.length();i++){            p_int[i] = p.charAt(i) - 'a';        }        int res = 0;        int maxLen = 0;        for( int i=0;i<p.length();i++ ){            if( i>0 && (p_int[i-1] + 1) % 26 == p_int[i]){                maxLen ++ ;            } else{                maxLen = 1;            }            count[p_int[i]] = Math.max(count[p_int[i]],maxLen);        }        for( int i=0;i<26;i++)            res += count[i];        return res;    }}