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 ""。 示例 1: 输入:strs = ["flower","flow","flight"] 输出:"fl" 示例 2: 输入:strs = ["dog","racecar","car"] 输出:"" 解释:输入不存在公共前缀。  
乘风破浪:LeetCode真题_014_Longest Common Prefix一、前言如何输出最长的共同前缀呢,在给定的字符串中,我们可以通过笨办法去遍历,直到其中某一个字符不相等了,这样就得到了最长的前缀。那么还有没有别的办法呢?二、Longest Common Prefix2.1 问题2.2 分析与解决由问题我们可以知道,所有的字符都是小写的,这样我们不...
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: strs = ["flower","flow","flight"] Output: "fl" Example 2: Input: strs = ["dog","racecar","car"] Output: "" Ex...
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...
public String longestCommonPrefix(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) {
给n个字符串,求最长公公前缀。直接逐位扫判断就行。 class Solution(object): def longestCommonPrefix(self, strs): """ :type strs: List[str] :rtype: str """ lenS = len(strs) if lenS == 0: return '' if lenS == 1: return strs[0] ...
1. public String longestCommonPrefix(String[] strs) { 2. if (strs.length == 0) { 3. return ""; 4. } 5. int miniLength = strs[0].length(); 6. for (String s : strs) { 7. if (s.length() < miniLength) { 8. miniLength = s.length(); ...
classSolution{public:stringlongestCommonPrefix(vector<string>&strs){if(strs.empty()){return"";}for(inti=0;i<strs[0].length();++i){for(intj=1;j<strs.size();++j){if(strs[j][i]!=strs[0][i]){returnstrs[0].substr(0,i);}}}returnstrs[0];}};...
# @Time : 2024/3/3 16:53 # @Author : fangel # @FileName : [leetcode] 14. 最长公共前缀.py # @Software : PyCharm class Solution:def longestCommonPrefix(self, strs: list[str]) -> str:# 步骤1：如果strs为单个元素，直接返回 if len(strs) == 1 or len(strs) == 0:r...
这道题真是有点emmm...每次提交都能发现自己的漏洞,查漏补缺做出来了。但是比较冗长,把代码和感想记录如下: class Solution: def longestCommonPrefix(self, strs): count = 0 if len(strs) > 1: if len(min(strs, key=len)) == 0: flag...