LeetCode - Longest Common Prefix

LeetCode - Longest Common Prefix

2013.12.1 23:23

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

Solution:

　　For two string s1 and s2, compare the elements from the beginning, when the end or the first mismatch is found, the common prefix is found.

　　For n strings s[n], perform (n - 1) times such comparings. If the current common prefix is already empty "", then the latter comparison is meaningless and cancelled.

　　Time complexity is O(n * max(strlen(s[i]))), that means the comparison can be performed at most (n - 1) times, and no single comparison will go longer than the longest string amongst s[n]. Space complexity is O(1).

　　Actually, if we find out the shortest string s_min amongst s[n], and start the first comparison from the shortest string. We'll gain a time complexity of O(n * min(strlen(s[i]))). Space complexity remains O(1).

　　Here is an example to show why order of the data matters:

　　　　['aaa', ''aaa', 'aaa', 'd']

　　　　['d', 'aaa', 'aaa', 'aaa']

　　　　The time needed for the two dataset above would be different, since order is different. It's better to put short ones on the front.

Accepted code:

 1 // 1RE, 1AC

 2 #include <algorithm>

 3 using namespace std;

 4 

 5 class Solution {

 6 public:

 7     string longestCommonPrefix(vector<string> &strs) {

 8         // IMPORTANT: Please reset any member data you declared, as

 9         // the same Solution instance will be reused for each test case.

10         int n = strs.size();

11         // 1RE here, didn't consider the case when $strs is empty.

12         if(n <= 0){

13             return "";

14         }

15         string res = strs[0];

16         

17         int i;

18         for(i = 1; i < n; ++i){

19             res = commonPrefix(res, strs[i]);

20             if(res == ""){

21                 break;

22             }

23         }

24         

25         return res;

26     }

27 private:

28     string commonPrefix(string s1, string s2) {

29         if(s1.length() > s2.length()){

30             return commonPrefix(s2, s1);

31         }

32         

33         int len1, len2;

34         

35         len1 = s1.length();

36         len2 = s2.length();

37         int i;

38         for(i = 0; i < len1; ++i){

39             if(s1[i] != s2[i]){

40                 break;

41             }

42         }

43         

44         return s1.substr(0, i);

45     }

46 };