这是一系列linkedin的面试题，比较接近于Indeed的面试题目，follow up加ＯＯ的多接口形式．
Shortest Word Distance
Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.
For example,
Assume that words =["practice", "makes", "perfect", "coding", "makes"]. Given word1 = “coding”, word2 = “practice”, return 3.
"makes", word2 = "coding", return 1. Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.这个是简单的扫描问题，two pointer解决，p1,p2分别记录word1和word2在wordlist当中的位置．一旦p1,p2有更新,则计算一次距离，和当前最小距离比较．这样，每次都是超前找最相近的另一个单词．可以找到word1附近离他最近的word2（有前右后）, word2附近离他最近的word1(有前右后）．总之可以保证找到最小距离．
时间复杂度Ｏ(n),空间复杂度Ｏ(1)
代码如下：
class Solution {public: int shortestDistance(vector & words, string word1, string word2) { int p1 = -1; int p2 = -1; int dis = words.size(); for(int i=0; i < words.size(); i++){ if (words[i] == word1) p1 = i; if (words[i] == word2) p2 = i; if (p1!=-1 && p2 != -1) dis = min(dis, abs(p1-p2)); } return dis; }}; Shortest Word Distance II
This is a follow up of . The only difference is now you are given the list of words and your method will be called repeatedly many times with different parameters. How would you optimize it?
Design a class which receives a list of words in the constructor, and implements a method that takes two words word1 and word2 and return the shortest distance between these two words in the list.
For example,
Assume that words =["practice", "makes", "perfect", "coding", "makes"]. Given word1 = “coding”, word2 = “practice”, return 3.
"makes", word2 = "coding", return 1. Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.Ｉ的follow up, 和Indeed的find k nearest stars 类似，注意到这里需要多次调用该方法，且参数不同．所以需要优化，不能每次都Ｏ(n)扫一遍，解决策略是 先建一个hashmap，把每个单词出现的index list先缓存起来，这样每次就是在word1的index list和word2的index list当中找距离最近的数字．其实就是而且因为hashmap按序添加，这里的两个list都是有序的．不需要预先排序.所以双指针一遍扫描，复杂度为两个list的长度相加Ｏ(n1+n2).代码如下： 核心思路为舍去不可能的方向．
class WordDistance(object): def __init__(self, words): """ initialize your data structure here. :type words: List[str] """ self.hash = {} for i in xrange(len(words)): if words[i] not in self.hash: self.hash[words[i]] = [i] else: self.hash[words[i]].append(i) def shortest(self, word1, word2): """ Adds a word into the data structure. :type word1: str :type word2: str :rtype: int """ #two pointer,min difference res = sys.maxint p1 = 0 p2 = 0 list1 = self.hash[word1] list2 = self.hash[word2] while p1 < len(list1) and p2 < len(list2): if list1[p1] >= list2[p2]: res = min(res, (list1[p1]-list2[p2])) p2 += 1 ＃此时增加P1,不会有更好的结果 else: res = min(res, (list2[p2] - list1[p1])) p1 += 1 ＃此时增加p2,b不会有更好的结果 return res Shortest Word Distance III
This is a follow up of . The only difference is now word1 could be the same as word2.
Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.
word1 and word2 may be the same and they represent two individual words in the list.
For example,
Assume that words =["practice", "makes", "perfect", "coding", "makes"]. Given word1 = “makes”, word2 = “coding”, return 1.
"makes", word2 = "makes", return 3. Note:
You may assume word1 and word2 are both in the list.主要是处理重复情况，思路是Ｉ的基础上加一个flag标志．代码如下：
class Solution(object): def shortestWordDistance(self, words, word1, word2): """ :type words: List[str] :type word1: str :type word2: str :rtype: int """ #word1 and word2 are the same words, how to handle the same situation into the one situation. res = sys.maxint if word1 == word2: flag = True else: flag = False p1 = -1 p2 = -1 for i in xrange(len(words)): if words[i] == word1: if flag and p1!= -1: res = min(res, i-p1) elif not flag and p2 != -1: res = min(res, i-p2) p1 = i if words[i] == word2: if not flag and p1 != -1: res = min(res, i-p1) p2 = i return res