import pandas as pd class Solution(object): def longestCommonPrefix(self, strs): for i in strs: if len(i) > 200 or len(i) < 1: # 控制列表strs长度 return False if not i.islower(): # 限制strs每一个元素必须为小写 return False strs_dict = {} #为strs的每一个元素创建一个空字...
使用python解这种题目时有非常多的trick.合理使用可以提速很多. 按行扫不使用trick版本,运行时间60ms.比较慢. classSolution(object):deflongestCommonPrefix(self, strs):""":type strs: List[str] :rtype: str"""ifnotstrs:return""forjinrange(len(strs[0])):foriinrange(1,len(strs)):ifj >= l...
1、当strs为空,直接输出“” 2、当strs中含有“”,直接输出“” 3、strs[0]的最长长度由最短公共长度l决定(code line:15) 1classSolution:2#@return a string3deflongestCommonPrefix(self, strs):4ifstrs ==[]:5return""6foriinrange(1,len(strs)):7l1 =len(strs[0])8l2 =len(strs[i])...
Memory Usage: 12.8 MB, less than 100.00% of Python3 online submissions classSolution:deflongestCommonPrefix(self,strs:List[str])->str:preStr=''ifnotlen(strs):# empty strsreturnpreStrforiinrange(len(strs[0])):firstLetter=strs[0][:i+1]flag=1forsinstrs:ifnots.startswith(firstLetter)...
class Solution { public: string longestCommonPrefix(vector<string> &strs) { if (strs.empty()) { return ""; } string common = strs[0]; vector<string>::size_type size = strs.size(); int length;//保存要比较的两个字符串的最小长度,只在最小长度范围内进行比较 ...
python # 方法一 时间O(mn),空间O(1) def longest_common_prefix(strs: [str]) -> str: """ 最长公共前缀 -> 纵向扫描 :param strs: [str] :return: str """ if not strs: return '' length, count = len(strs[0]), len(strs) ...
//Divide-and-Conquer Approach, python, 44ms class Solution { public: string longestCommonPrefix(vector<string>& strs) { if (strs.empty()) return ""; for (int pos = 0; pos < strs[0].length(); pos++) for (int i = 1; i < strs.size(); i++) ...
classSolution{publicStringlongestCommonPrefix(String[]strs){if(strs==null||strs.length==0)return"";String prefix=strs[0];for(int i=1;i<strs.length;i++){while(strs[i].indexOf(prefix)!=0){prefix=prefix.substring(0,prefix.length()-1);if(prefix.isEmpty())return"";}}returnprefix;...
class Solution: def longgestCommonPrefix(self,strs): if len(strs)<1 or len(strs)>200: return None #先获取列表最短字符串长度,即为最大公共前缀的长度上限;同时获取此长度中排最靠前的字符串记录下来 lenth = 100000000000 for elem in strs: if len(elem)< lenth: lenth = len(elem) a = el...
Python 3 实现: 源代码已上传Github,持续更新。 """ 14. Longest Common Prefix Write a function to find the longest common prefix string amongst an array of strings. """classSolution:deflongestCommonPrefix(self,strs):""" :type strs: List[str] ...