Trie and Suffix

前缀树(trie)和后缀树(suffix)是两种非常重要的数据结构,往往用来加速字符串的匹配和查找。

具体的介绍可见下面的资料:

http://blog.youkuaiyun.com/luowen3405/archive/2011/04/01/6295839.aspx

http://blog.youkuaiyun.com/TsengYuen/archive/2009/11/16/4815921.aspx

 

Trie和Suffix最大的不同是,Trie是面向一个字符串的集合,而Suffix是面向一个字符串

 

在面试题目中,一些经典的前缀和后缀应用如下:

Suffix Tree:

求两个字符串的最长公共子串。

1. The longest common substring of two strings, txt1 and txt2, can be found by building a generalized suffix tree for txt1 and txt2: Each node is marked to indicate if it represents a suffix of txt1 or txt2 or both. The 

deepest node marked for both txt1 and txt2 represents the longest common substring.

2. Equivalently, one can build a (basic) suffix tree for the string txt1$txt2#, where `$' is a special terminator for txt1 and `#' is a special terminator for txt2. The longest common substring is indicated by the deepest fork node that has both `...$...' and `...#...' (no $) beneath it.

求一个字符串的最长重复子串
       Add a special end of string character, e.g. `$', to txt[1..n] and build a suffix tree; the longest repeated substring of txt[1..n] is indicated by the deepest fork node in the suffix tree, where depth is measured by the number of characters traversed from the root, i.e., `issi' in the case of `mississippi'. The longest repeated substring can be found in O(n) time using a suffix tree.

      求一个字符串的最长回文
        A palindrome is a string, P, such that P=reverse(P). e.g. `abba'=reverse(`abba'). e.g. `ississi' is the longest palindrome in `mississippi'. The longest palindrome of txt[1..n] can be found in O(n) time, e.g. by building the suffix tree for txt$reverse(txt)# or by building the generalized suffix tree for txt and reverse(txt).

 

Trie

      字典
        A common application of a trie is storing a dictionary, such as one found on a mobile telephone. Such applications take advantage of a trie's ability to quickly search for, insert, and delete entries; however, if 
storing dictionary words is all that is required (i.e. storage of information auxiliary to each word is not required), a minimal acyclic deterministic finite automaton would use less space than a trie.

D. Good Substrings time limit per test2 seconds memory limit per test512 megabytes You've got string s, consisting of small English letters. Some of the English letters are good, the rest are bad. A substring s[l...r] (1 ≤ l ≤ r ≤ |s|) of string s  =  s1s2...s|s| (where |s| is the length of string s) is string  slsl + 1...sr. The substring s[l...r] is good, if among the letters  sl, sl + 1, ..., sr there are at most k bad ones (look at the sample's explanation to understand it more clear). Your task is to find the number of distinct good substrings of the given string s. Two substrings s[x...y] and s[p...q] are considered distinct if their content is different, i.e. s[x...y] ≠ s[p...q]. Input The first line of the input is the non-empty string s, consisting of small English letters, the string's length is at most 1500 characters. The second line of the input is the string of characters "0" and "1", the length is exactly 26 characters. If the i-th character of this string equals "1", then the i-th English letter is good, otherwise it's bad. That is, the first character of this string corresponds to letter "a", the second one corresponds to letter "b" and so on. The third line of the input consists a single integer k (0 ≤ k ≤ |s|) — the maximum acceptable number of bad characters in a good substring. Output Print a single integer — the number of distinct good substrings of string s. Examples InputCopy ababab 01000000000000000000000000 1 OutputCopy 5 InputCopy acbacbacaa 00000000000000000000000000 2 OutputCopy 8 Note In the first example there are following good substrings: "a", "ab", "b", "ba", "bab". In the second example there are following good substrings: "a", "aa", "ac", "b", "ba", "c", "ca", "cb".
03-28
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值