题目链接:https://leetcode-cn.com/problems/longest-common-prefix/
题目描述:
如果不存在公共前缀，返回空字符串 ""。
示例:
示例 1:
输入: ["flower","flow","flight"]
输出: "fl"
示例 2:
输入: ["dog","racecar","car"]
输出: ""
解释: 输入不存在公共前缀。
说明:
所有输入只包含小写字母 a-z 。
思路:
思路1:
python特性,取每一个单词的同一位置的字母,看是否相同.
思路2:
取一个单词s,和后面单词比较,看s与每个单词相同的最长前缀是多少!遍历所有单词
思路3:
按字典排序数组,比较第一个,和最后一个单词,有多少前缀相同.
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代码:
python
思路一:
class Solution:
def longestCommonPrefix(self, strs):
"""
:type strs: List[str]
:rtype: str
"""
res = ""
for tmp in zip(*strs):
tmp_set = set(tmp)
if len(tmp_set) == 1:
res += tmp[0]
else:
break
return res
思路二:
python
class Solution:
def longestCommonPrefix(self, s: List[str]) -> str:
if not s:
return ""
res = s[0]
i = 1
while i < len(s):
while s[i].find(res) != 0:
res = res[0:len(res)-1]
i += 1
return res
java
class Solution {
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) return "";
String res = strs[0];
int i = 1;
while (i < strs.length) {
while (strs[i].indexOf(res) != 0) {
res = res.substring(0, res.length() - 1);
}
i += 1;
}
return res;
}
}
思路三:
python
class Solution:
def longestCommonPrefix(self, s: List[str]) -> str:
if not s:
return ""
s.sort()
n = len(s)
a = s[0]
b = s[n-1]
res = ""
for i in range(len(a)):
if i < len(b) and a[i] == b[i]:
res += a[i]
else:
break
return res
java
class Solution {
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) return "";
StringBuilder res = new StringBuilder();
Arrays.sort(strs);
// 字符串转数组
char[] a = strs[0].toCharArray();
char[] b = strs[strs.length - 1].toCharArray();
for (int i = 0; i < a.length; i++) {
if (i < b.length && a[i] == b[i]) {
res.append(a[i]);
}
else{
break;
}
}
return res.toString();
}
}
