标签：point param ping sam none ret modal tran lex
给出一个单词列表和两个单词单词1,单词2，返回列表中这两个单词之间的最短距离。
样例 1:
输入：["practice", "makes", "perfect", "coding", "makes"],"coding","practice"
输出：3
解释：index("coding") - index("practice") = 3
样例 2:
输入：["practice", "makes", "perfect", "coding", "makes"],"makes","coding"
输出：1
解释：index("makes") - index("coding") = 1
您可以假定单词1不等于单词2，而单词1和单词2在列表中都存在。
class Solution: """ @param words: a list of words @param word1: a string @param word2: a string @return: the shortest distance between word1 and word2 in the list """ ‘‘‘ 1.word1和word2可能都存在多个的情况，所以需要两层循环，取出所有可能的距离。(取出列表全部重复元素的索引) 2.初始化res，将两个单词距离全部append到res里面，返回最小值 ‘‘‘ def shortestDistance(self,words,word1,word2): res = [] l_word1 = [index for index,value in enumerate(words) if value == word1] l_word2 = [index for index,value in enumerate(words) if value == word2] #此时求出全部的距离，append到res里面 for i in l_word1: for j in l_word2: s = abs(i - j) res.append(s) return min(res)
标签：point param ping sam none ret modal tran lex
原文地址：https://www.cnblogs.com/yunxintryyoubest/p/12590515.html
