LeetCode 14: Longest Common Prefix-优快云博客

本文介绍了一种使用二分划分法寻找字符串数组中最长公共前缀的方法。该算法的时间复杂度为O(mlogn)，其中n为字符串数量，m为字符串长度。通过递归地将字符串数组划分为更小的部分来寻找公共前缀。

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Difficulty: 2

Frequency: 1

Problem:

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

Solution:

class Solution {
public:
    string longestCommonPrefix(vector<string> &strs) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (strs.size()==0)
        {   
            return string("");
        }
        return longestCommonPrefixBinaryDivide(strs, 0, strs.size()-1);
    }
    string longestCommonPrefixBinaryDivide(vector<string> & strs, int i_begin, int i_end)
    {
        if (i_begin==i_end)
            return strs[i_begin];
        
        if (i_begin==(i_end - 1))
        {
            int i = 0;
            while (i<strs[i_begin].size()&&i<strs[i_end].size()&&strs[i_begin][i]==strs[i_end][i])
            {
                ++i;
            }
            return string(strs[i_begin].substr(0, i));
        }
        
        string str1(longestCommonPrefixBinaryDivide(strs, i_begin, (i_begin+i_end)/2));
        string str2(longestCommonPrefixBinaryDivide(strs, (i_begin+i_end)/2+1, i_end));
        int i = 0;
        while (i<str1.size()&&i<str2.size()&&str1[i]==str2[i])
        {
            ++i;
        }
//        string answer(str1.substr(0, i));
        return string(str1.substr(0, i));
    }
};

Note:

It is like a binary search. Time complexity is O(mlogn). n is the number of strings, m is the strings' length.

Don't know if there is better answer. If you know please tell me.