Leetcode-467. Unique Substrings in Wraparound String

最新推荐文章于 2020-07-11 17:37:46 发布

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最新推荐文章于 2020-07-11 17:37:46 发布

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分类专栏：算法文章标签： java leetcode 算法

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本文介绍了一种高效算法，用于计算一个特定字符串在无限循环的字母表字符串中不同非空子串的数量。该算法特别适用于处理长度可能超过10,000的大字符串。

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发优快云，mcf171专栏。

博客链接：mcf171的博客

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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.

看完论坛里关于这个题的解法，我只能说服了，和当初看到异或操作感觉是完全一样的感觉。。。

public class Solution {
    public int findSubstringInWraproundString(String p) {
        // count[i] is the maximum unique substring end with ith letter.
        // 0 - 'a', 1 - 'b', ..., 25 - 'z'.
        int[] count = new int[26];
        
        // store longest contiguous substring ends at current position.
        int maxLengthCur = 0; 

        for (int i = 0; i < p.length(); i++) {
            if (i > 0 && (p.charAt(i) - p.charAt(i - 1) == 1 || (p.charAt(i - 1) - p.charAt(i) == 25))) {
                maxLengthCur++;
            }
            else {
                maxLengthCur = 1;
            }
            
            int index = p.charAt(i) - 'a';
            count[index] = Math.max(count[index], maxLengthCur);
        }
        
        // Sum to get result
        int sum = 0;
        for (int i = 0; i < 26; i++) {
            sum += count[i];
        }
        return sum;
    }
}