# Leetcode 14:Longest Common Prefix 最长公共前缀

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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 "".

编写一个函数来查找字符串数组中的最长公共前缀。

如果不存在公共前缀,返回空字符串 ""

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.

Note:

All given inputs are in lowercase letters a-z.

说明:

所有输入只包含小写字母 a-z

解题思路Java:

​ 很简单又很经典的一道题,我的思路起先是 把第字符串组第一个字符串转为char型。利用StringBuilder逐一累加相同字符。由于字符串长度不一,可以先遍历找出最小长度字符串,这里我选择抛错的形式,减少一次遍历。

代码:

class Solution {
    public String longestCommonPrefix(String[] strs) {
        int strLen=strs.length;
        if(strLen==0) return "";//空字符串组返回""
        char[] temp=strs[0].toCharArray();
        StringBuilder str = new StringBuilder();
        for (int i=0;i<strs[0].length();i++){//以第一个字符串长度开始比较
            for (int j=1;j<strLen;j++){
                try {
                    if(temp[i]!=strs[j].charAt(i)){
                        return str.toString();
                    }
                }catch (IndexOutOfBoundsException e){//抛出错误,这里错误是指索引超出字符串长度
                    return strs[j];
                }
            }
            str.append(temp[i]);
        }
        return strs[0];
    }
}

​ 后面想到Java有 subString() 方法,可指定长度截取字符串,无需转为 char[] 型,但是在 Leetcode 提交时反而不如上面这种方式运算快,这也说明了Java不支持运算符重载,使用 substring() 每次新建一个String字符串,效率并不高。

​ 最后看到一个方法,大致思路是找到最小长度字符串,从大到小截取字符串,既然用到 subString() 方法,不如就从后向前,因为题目是找出最长公众前缀,从大到小效率很高。具体请看:

 public class Solution {
    public String longestCommonPrefix(String[] strs) {
        if(strs.length==0) return "";
		int min=Integer.MAX_VALUE;
		String minStr="";
		for(int i=0;i<strs.length;i++){//找出最小长度字符串
			if(min>strs[i].length()){
				minStr=strs[i];
				min=strs[i].length();
			}
		}
		if(min==0) return "";
		for(int i=min;i>=0;i--){//最小长度字符串从长到短截取
			String standard=minStr.substring(0, i);
			int j=0;
			for(j=0;j<strs.length;j++){
				if(strs[j].substring(0, i).equals(standard)) continue;
				else break;
			}
			if(j==strs.length) return standard;
		}
		return "";
    }
}

原代码链接: blog.csdn.net/qq_14927217…

解题思路py3:

​ 再次投机取巧,os.path 封装函数 commonprefix() 一步到位。

代码:

class Solution(object):
    def longestCommonPrefix(self, strs):
        import os
        return os.path.commonprefix(strs)

​ 其实该函数是利用ASCll码比较的特性来编写的,源码:

def commonprefix(m):
    "Given a list of pathnames, returns the longest common leading component"
    if not m: return ''
    # Some people pass in a list of pathname parts to operate in an OS-agnostic
    # fashion; don't try to translate in that case as that's an abuse of the
    # API and they are already doing what they need to be OS-agnostic and so
    # they most likely won't be using an os.PathLike object in the sublists.
    if not isinstance(m[0], (list, tuple)):
        m = tuple(map(os.fspath, m))
    s1 = min(m)
    s2 = max(m)
    for i, c in enumerate(s1)://枚举得到s1的每一个字符及其索引
        if c != s2[i]:
            return s1[:i]
    return s1

尽管如此,py3这段代码的执行速度依然远比Java慢的多。

**注:**ASCll码比较大小并非是按照所有字符的ASCll累加之和比较,是从一个字符串第一个字符开始比较大小,如果不相同直接得出大小结果,后面的字符不在比较。