[leetcode] 243. Shortest Word Distance 解题报告

题目链接:https://leetcode.com/problems/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.
Given word1 = "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.

思路: 需要注意的是可能两个单词都会在字典里出现很多次, 因此我们可以用hash table保存一个单词的多个位置, 然后枚举最小的距离.在找最小距离的时候可以利用一个小小的技巧将时间复杂度由O(m*n)降为O(m+n).
代码如下:

class Solution {public:    int shortestDistance(vector<string>& words, string word1, string word2) {        unordered_map<string, vector<int>> hash;        for(int i=0; i<words.size(); i++) hash[words[i]].push_back(i);        int i = 0, j =0, ans = INT_MAX;        while(i < hash[word1].size() && j < hash[word2].size())        {            ans = min(ans, abs(hash[word1][i]-hash[word2][j]));            hash[word1][i]<hash[word2][j]?i++:j++;        }        return ans;    }};

python

class Solution(object):    def shortestDistance(self, words, word1, word2):        """        :type words: List[str]        :type word1: str        :type word2: str        :rtype: int        """        w1 = [i for i in range(len(words)) if words[i]==word1]        w2 = [i for i in range(len(words)) if words[i]==word2]        return min([abs(i-j) for i in w1 for j in w2])