467. Unique Substrings in Wraparound String

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.

思路：

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。

class Solution {
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.char
4000
At(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;
}
}