题解 | #最长公共前缀#

#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param strs string字符串一维数组 
# @return string字符串
#
class Solution:
    def longestCommonPrefix(self , strs: List[str]) -> str:
        # write code here
        n = len(strs)
        if n == 0:
            return ""
            # 处理空数组
        short = len(strs[0])
        p = 0
        for i in range(n):
            if len(strs[i]) < short:
                short = len(strs[i])
                p = i
                # 找出最短的字符串并标记位置
        for i in range(short):
            # 遍历最短的字符串,最长子串不会超过该字符串
            bs = strs[p][i]
            # 从首字符开始比对
            for j in range(n):
                if strs[j][i] != bs:
                    # 遇到不相等的字符，说明之前的字符都匹配成功
                    return strs[p][:i]
        return strs[p]