[LeetCode] 244. Shortest Word Distance II 最短单词距离 II

2021年09月15日 阅读数:1
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This is a follow up of Shortest Word Distance. 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?html

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.python

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].app

Given word1 = “coding”word2 = “practice”, return 3.
Given word1 = "makes"word2 = "coding", return 1.函数

Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.post

243. Shortest Word Distance 的拓展,不一样的是此次须要屡次调用求最短单词距离的函数。url

Python:code

# Time:  init: O(n), lookup: O(a + b), a, b is occurences of word1, word2
# Space: O(n)
import collections

class WordDistance:
    # initialize your data structure here.
    # @param {string[]} words
    def __init__(self, words):
        self.wordIndex = collections.defaultdict(list)
        for i in xrange(len(words)):
            self.wordIndex[words[i]].append(i)

    # @param {string} word1
    # @param {string} word2
    # @return {integer}
    # Adds a word into the data structure.
    def shortest(self, word1, word2):
        indexes1 = self.wordIndex[word1]
        indexes2 = self.wordIndex[word2]

        i, j, dist = 0, 0, float("inf")
        while i < len(indexes1) and j < len(indexes2):
            dist = min(dist, abs(indexes1[i] - indexes2[j]))
            if indexes1[i] < indexes2[j]:
                i += 1
            else:
                j += 1

        return dist  

C++:htm

class WordDistance {
public:
    WordDistance(vector<string>& words) {
        for (int i = 0; i < words.size(); ++i) {
            m[words[i]].push_back(i);
        }
    }

    int shortest(string word1, string word2) {
        int res = INT_MAX;
        for (int i = 0; i < m[word1].size(); ++i) {
            for (int j = 0; j < m[word2].size(); ++j) {
                res = min(res, abs(m[word1][i] - m[word2][j]));
            }
        }
        return res;
    }
    
private:
    unordered_map<string, vector<int> > m;
};

C++:blog

class WordDistance {
public:
    WordDistance(vector<string>& words) {
        for (int i = 0; i < words.size(); ++i) {
            m[words[i]].push_back(i);
        }
    }

    int shortest(string word1, string word2) {
        int i = 0, j = 0, res = INT_MAX;
        while (i < m[word1].size() && j < m[word2].size()) {
            res = min(res, abs(m[word1][i] - m[word2][j]));
            m[word1][i] < m[word2][j] ? ++i : ++j;
        }
        return res;
    }
    
private:
    unordered_map<string, vector<int> > m;
};

  

  

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[LeetCode] 243. Shortest Word Distance 最短单词距离

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