[LeetCode]#14 Longest Common Prefix

一、题目

Write a function to find the longest common prefix string amongst an array of strings.

找到一组str的最长前缀

 

二、解析

这题比较水，只需比较第1、2个str，将相同的前缀记录下来，并拿这个去与第3个str做比较。如果前缀可以融入第三个str，则说明第三个str也有相同前缀，就往下找第4个str。如果不能融入，那就把前缀最后一位去掉，再去判断是否可以融入第三个str，如果不行再去最后一位，如此循环。直到前缀被去完，或者前缀可以融入所有的str为之。还是很好理解的。

 

三、代码

class Solution:
    # @param {string[]} strs
    # @return {string}
    def longestCommonPrefix(self, strs):
        #in thie problem, we need to find the longest common prefix
        #i set the result string <str>longest to store the common prefix
        #first I let longest = strs[0], and to compare each strs[i], do "-" operation
        #eg:strs=['she', "shanghai", "super"]
        #round0: longest = "she", len=3
        #round1: "she" != "shanghai"[:3], so cut the last one in longest
        #        longest = "sh", len=2
        #        "sh" == "shanghai"[:2],ok ,the next
        #round2: "sh" != "shper"[:2], cut the last one in longest
        #        longest = "s", len=1
        #        "s" == super[:1], ok, return longest = "s"
        length = len(strs)
        if length == 0:
            return ""
        elif length == 1:
            return strs[0]
        else:
            longest = strs[0]
            for i in range(1, length):
                if strs[i] == "":
                    return ""
                else:
                    while longest != strs[i][:len(longest)]:
                        len_longest = len(longest)
                        longest = longest[:len_longest - 1]
                        if longest == "":
                            return ""
            return longest

 

四、总结

这是两周前A的题了吧，最近一直比较忙，发生了一些变故，没怎么刷题了。其实也不是说没时间，而是有时候累的脑子都不会转了。。还是要多多练手，继续保持节奏。