2018-09-01 22:50:59
问题描述:
问题求解:
如果单纯的遍历判断,那么如何去重保证unique是一个很困难的事情,事实上最初我就困在了这个点上。
后来发现是一个动态规划的问题,可以将每个字符结尾的最长长度进行保存,这样就巧妙的解决的重复的问题。
- 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 ins
sinces
has infinite length.- Now we know the max number of unique substrings in
p
ends with‘a‘, ‘b‘, ..., ‘z‘
and those substrings are all ins
. Summary is the answer, according to the question.
public int findSubstringInWraproundString(String p) { int res = 0; int[] dp = new int[26]; int maxLen = 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)) { maxLen++; } else maxLen = 1; dp[p.charAt(i) - ‘a‘] = Math.max(dp[p.charAt(i) - ‘a‘], maxLen); } for (int i = 0; i < 26; i++) res += dp[i]; return res; }
原文地址:https://www.cnblogs.com/TIMHY/p/9572041.html
时间: 2024-10-20 12:06:59