class Solution:
    def longestCommonPrefix(self, strs):
        longest=""
        for i in range(min([len(s) for s in strs])):
            yn=[]
            for s2 in strs:
                yn.append(s2.startswith(strs[0][:i+1]))
            if not (False in yn):
                longest=strs[0][:i+1]
        return longest

解题思路：

遍历的次数由最短的字符串决定

判断每个字符串是否有同样的前缀

是则把最长前缀设置为当前前缀

最后返回即可

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