[LeetCode]#14 Longest Common Prefix

一、题目

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

 

二、解析

题目是要找到一个str的list中，所有元素最长的公共前缀。例如she, shanghai, s的公共前缀是s。

我的想法是，既然找公共前缀(longest)，那么它在list所有元素中是肯定存在的，并且是在前面的。所以，我们只需比较longst是否等于元素的前[:len(longest)]即可。、如果相等，那么就说明longest也是当前元素的公共前缀；如果不相等，说明实际的前缀是短的，就将longest最后一个字母删除，再进行比较。直到与当前元素相同，或者longest变为空为止。

挺简单的。

 

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

在思考解法的时候，又有一段时间陷入了循环空想。静下心来，一点一点分析，自己不糊弄自己，想清楚细节，这样很快就跳出来，理清了思路。A了这道题到没有很开心，这个跳出来的过程我倒是很享受，哈哈~~