LeetCode 467 Unique Substrings in Wraparound String (思维题)

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.

题目分析：开始一直纠结于如何枚举全部子串，中间存在着容斥问题，很快发现想偏了，如果可以回避重叠带来的容斥就好做了，最后想到用后缀的方式，枚举分别以26个字母结尾的连续子串的最大串长，它们的和就是答案。为什么这就是答案？首先最大串长实际表示的是该串后缀子串的个数，举个例子具体分析：abcbcdcde

以a，b，c，d，e结尾的最长连续子串分别是a，ab，abc，bcd，cde，因此答案是12，首先四个子串的最后一个字符都是不同的，因此它们的所有后缀子串肯定也都不同，现在只需要证明不遗漏即可，直接证不好证，尝试反证，假设有遗漏，遗漏的那个串为s且末位为‘?’，那么‘?’肯定是a，b，c，d，e中的某个，又因为我们取的答案子串是以a，b，c，d，e结尾的最长的连续子串，因此s必然是那些后缀串中的某一个。

public class Solution {    public int findSubstringInWraproundString(String p) {        p += "#";        int len = p.length(), ans = 0, tmp = 1;        int[] dp = new int[26];        Arrays.fill(dp, 0);        for (int i = 0; i < len - 1; i ++) {            char cur = p.charAt(i);            if ((cur - 'a' + 1) % 26 == p.charAt(i + 1) - 'a') {                tmp ++;            } else {                tmp = 1;            }            dp[cur - 'a'] = Math.max(dp[cur - 'a'], tmp);         }        for (int i = 0; i < 26; i ++) {            ans += dp[i];        }        return ans;    }}