[LeetCode] 14. Longest Common Prefix 最长共同前缀

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

If there is no common prefix, return an empty string "".

Example 1:

Input: ["flower","flow","flight"]
Output: "fl"

Example 2:

Input: ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.

这题有好几种解法，个人认为会1，2的解法就可以了，但这种多方法解题的思路可以好好学习一下。具体可参考：Longest Common Prefix

1. 一个一个字符串取比较word by word matching：

先拿前2个，从第一位开始比较，直到发现有不同的字符，此时前面一样的字符串在去和后面的字符串比较，直到结束。可以用递归。

Time: O(n*m) (n是字符串个数，m是字符串最长长度)  Space: O(m)

2. 一个字符一个字符的比较character by character matching：

所有的字符串同时比较第1个，第2个.......，发现有不同的出现，之前一样的就是找到的最长共同前缀。Time: O(n*m) (n是字符串个数，m是字符串最长长度) Space: O(m)

3. divide and conquer: 

把所有字符串分成两组，分别去比较，最后都剩一个的时候，两组在比较。Time： O(n*m), Space: O(n*logm)

4. 二分法Binary Search：

先找到最短的字符串，然后把这个最短的字符串二分成前面和后面两部分，前面的和所有剩下字符串比较，如果一样在比较后面的，如果有不一样的，则后面的部分不用比较了，前面的部分在二分比较。Time： O(n*m*logm), Space: O(m)

5. 使用Trie:

首先了解Trie数据结构，然后把所有的字符串都执行一遍插入到Trie，然后读取Trie中最后一个没有分支的node，此时之前这些字符就是答案。

Time： O(n*m + m), Space: O(26*m*n) ~ O(m*n)

Java: Method 1

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public class Solution {     public String longestCommonPrefix(String[] strs) {         if (strs == null || strs.length == 0) {             return "";         }         String prefix = strs[0];         for(int i = 1; i < strs.length; i++) {             int j = 0;             while( j < strs[i].length() && j < prefix.length() && strs[i].charAt(j) == prefix.charAt(j)) {                 j++;             }             if( j == 0) {                 return "";         }             prefix = prefix.substring(0, j);         }         return prefix;     }   }

Java: Method 2

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public class Solution {     public String longestCommonPrefix(String[] strs) {         if (strs == null || strs.length == 0) return "";         String res = new String();         for (int j = 0; j < strs[0].length(); ++j) {             char c = strs[0].charAt(j);             for (int i = 1; i < strs.length; ++i) {                 if (j >= strs[i].length() || strs[i].charAt(j) != c) {                     return res;                 }             }             res += Character.toString(c);         }         return res;     } }

Java: Method 2

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public class Solution {     public String longestCommonPrefix(String[] strs) {         if (strs == null || strs.length == 0) return "";         for (int j = 0; j < strs[0].length(); ++j) {             for (int i = 0; i < strs.length - 1; ++i) {                 if (j >= strs[i].length() || j >= strs[i + 1].length() || strs[i].charAt(j) != strs[i + 1].charAt(j)) {                     return strs[i].substring(0, j);                   }             }         }         return strs[0];     } }

Python: Method 2

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class Solution(object):     def longestCommonPrefix(self, strs):         """         :type strs: List[str]         :rtype: str         """         if not strs:             return ""           for i in xrange(len(strs[0])):             for string in strs[1:]:                 if i >= len(string) or string[i] != strs[0][i]:                     return strs[0][:i]         return strs[0]

C++: Method 2

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class Solution { public:        string longestCommonPrefix(vector<string> &strs) {         if (strs.size() == 0) {             return "";         }                   string prefix = "";         for (int i = 0; i < strs[0].length(); i++) {             for (int j = 1; j < strs.size(); j++) {                 if (strs[j][i] != strs[0][i]) {                     return prefix;                 }             }             prefix += strs[0][i];         }                   return prefix;     } };

　 

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