題目

https://leetcode.com/problems/unique-substrings-in-wraparound-string/

題解

1、dp 超時版本

class Solution {public int findSubstringInWraproundString(String p) {HashSet<Integer>[] arr = new HashSet[26]; // <結尾字母，長度>for (int i = 0; i < 26; i++) {arr[i] = new HashSet();}arr[p.charAt(0) - 'a'].add(1);int curLen = 1; // 避免誤將斷鏈情況計入結果for (int i = 1; i < p.length(); i++) {arr[p.charAt(i) - 'a'].add(1);if (p.charAt(i) == 'a' && p.charAt(i - 1) == 'z') {curLen++;for (int len : arr[25]) {if (len + 1 <= curLen) arr[0].add(len + 1);}} else if (p.charAt(i - 1) == p.charAt(i) - 1) {curLen++;for (int len : arr[p.charAt(i - 1) - 'a']) {if (len + 1 <= curLen) arr[p.charAt(i) - 'a'].add(len + 1);}} else {curLen = 1;}}int result = 0;for (int i = 0; i < 26; i++) {result += arr[i].size();}return result;} }

2、dp 答案版本

參考：Concise Java solution using DP

After failed with pure math solution and time out with DFS solution, I finally realized that this is a DP problem…
The idea is, if we know the max number of unique substrings in p ends with 'a', 'b', ..., 'z', then the summary of them is the answer. Why is that?

The max number of unique substring ends with a letter equals to the length of max contiguous substring ends with that letter. Example "abcd", the max number of unique substring ends with 'd' is 4, apparently they are "abcd", "bcd", "cd" and "d".

If there are overlapping, we only need to consider the longest one because it covers all the possible substrings. Example: "abcdbcd", the max number of unique substring ends with 'd' is 4 and all substrings formed by the 2nd "bcd" part are covered in the 4 substrings already.

No matter how long is a contiguous substring in p, it is in s since s has infinite length.

Now we know the max number of unique substrings in p ends with 'a', 'b', ..., 'z' and those substrings are all in s. Summary is the answer, according to the question.

Hope I made myself clear…

class Solution {public int findSubstringInWraproundString(String p) {int[] dp = new int[26]; // <結尾字母，以當前字母結尾的最長串長度>dp[p.charAt(0) - 'a'] = 1;int curLen = 1;for (int i = 1; i < p.length(); i++) {if (p.charAt(i - 1) == p.charAt(i) - 1 || p.charAt(i) == 'a' && p.charAt(i - 1) == 'z') {curLen++;} else {curLen = 1;}dp[p.charAt(i) - 'a'] = Math.max(dp[p.charAt(i) - 'a'], curLen);}int result = 0;for (int i = 0; i < 26; i++) {result += dp[i];}return result;} }

總結

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