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Problem Statement You are given a string input. You are to find the Longest substring of input such that the reversal of the substring is also a substring of input. In case of a tie, return the string that occurs earliest in input. Definition Class: ReverseSubstring Method: findReversed Parameters: string Returns: string Method signature: string findReversed(string input) (be sure your method is public) Notes The substring and its reversal may overlap partially or completely. The entire original string is itself a valid substring (see example 4). Constraints input will contain between 1 and 50 characters, inclusive. Each character of input will be an uppercase letter ( A - Z ). Examples 0) "XBCDEFYWFEDCBZ" Returns: "BCDEF" We see that the reverse of BCDEF is FEDCB, which appears later in the string. 1)
上傳時間： 2015-09-21
上傳用戶：sunjet
LCS Algorithm, this is a c++ code for lcs(Longest Common Subsequence)
標簽： Subsequence Algorithm Longest Common
上傳時間： 2013-12-25
上傳用戶：李夢晗
Longest Ordered Subsequence,acm必備習題
標簽： Subsequence Longest Ordered acm
上傳時間： 2014-10-27
上傳用戶：1159797854
Problem B:Longest Ordered Subsequence A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All Longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
標簽： Subsequence sequence Problem Longest
上傳時間： 2016-12-08
上傳用戶：busterman
最長子序列,英文版,電子書 ACM文章1977+Algorithms+for+the+Longest+Common+Subsequence+Problem
標簽： Subsequence Algorithms Longest Problem
上傳時間： 2014-01-27
上傳用戶：zhuimenghuadie
Instead of finding the Longest common subsequence, let us try to determine the length of the LCS. Then tracking back to find the LCS. Consider a1a2…am and b1b2…bn. Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn Let Li j denote the length of the Longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
標簽： the subsequence determine Instead
上傳時間： 2013-12-17
上傳用戶：evil
This scheme is initiated by Ziv and Lempel [1]. A slightly modified version is described by Storer and Szymanski [2]. An implementation using a binary tree is proposed by Bell [3]. The algorithm is quite simple: Keep a ring buffer, which initially contains "space" characters only. Read several letters from the file to the buffer. Then search the buffer for the Longest string that matches the letters just read, and send its length and position in the buffer.
標簽： initiated described modified slightly
上傳時間： 2014-01-09
上傳用戶：sk5201314
參考算法導論寫的LCS算法，仿照STL的泛型風格，適用于多種STL容器中的各種類型數據構成的序列的最大公共子序列（Longest Common Subsequence）問題求解。
上傳時間： 2014-11-22
上傳用戶：stvnash
這是動態規劃中，求最長公共子序列（Longest common string）的源代碼。自己編寫執行。程序簡單，有注釋。
標簽： 動態規劃
上傳時間： 2013-12-29
上傳用戶：xuanjie
This program sets up EVA Timer 1, EVA Timer 2, EVB Timer 3 and EVB Timer 4 to fire an interrupt on a period overflow. A count is kept each time each interrupt passes through the interrupt service routine. EVA Timer 1 has the shortest period while EVB Timer4 has the Longest period.
上傳時間： 2013-12-25
上傳用戶：康郎
