Leetcode: longest common prefix

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Write a function to find the longest common prefix string amongst an array of strings.

The meaning of the question is not very clear, and it is understood as the elders of any two string prefixes. However, this question is to find the longest common prefix of all strings, that is, all strings in the array contain this prefix.

Algorithm 1: Character-by-character comparison. The time complexity is O (n * l). N indicates the number of strings and L indicates the length of the longest prefix.

Class solution {public: String longestcommonprefix (vector <string> & STRs) {int n = STRs. size (); string res; If (n = 0) return res; For (INT Pos = 0; POS <STRs [0]. size (); POS ++) // The maximum prefix length cannot exceed STRs [0]. size (), character-by-character comparison {for (int K = 1; k <n; k ++) // The POS character of STRs [0] And STRs [1... n-1 if (STRs [K]. size () = POS | STRs [k] [POS]! = STRs [0] [POS]) return res;} res. push_back (STRs [0] [POS]) ;}return res ;}};

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Algorithm 2: 0th strings and other strings are prefixed one by one. The longest prefix length of all strings = min {prefixlength (STRs [0], STRs [I]), 0 <I <n} address of this Article

Class solution {public: String longestcommonprefix (vector <string> & STRs) {int n = STRs. size (); If (n = 0) Return ""; int Len = STRs [0]. size (); // The length of the longest prefix for (INT I = 1; I <n; I ++) {int K; For (k = 0; k <min (Len, (INT) STRs [I]. size (); k ++) if (STRs [0] [k]! = STRs [I] [k]) break; If (LEN> K) Len = K;} return STRs [0]. substr (0, Len );}};

 

The time complexity of algorithm 2 is O (n * m), n is the number of strings, M = STRs [0] and the average length of the prefix of other strings, the number of characters in algorithm 2 is more than that in algorithm 1. Example:

 

STRs [0]: abcdef

STRs [1: ABCDE

STRs [2: abcdk

STRs [3] AB

According to algorithm 1, only the first three characters of each string need to be compared; according to algorithm 2, STRs [0] and the other three strings need to be compared 6, 5, 3 Characters

 

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