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?
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.
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.
243. Shortest Word Distance 的拓展，不同的是这次需要多次调用求最短单词距离的函数。
Python:
# 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++:
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++:
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;
};
类似题目：
[LeetCode] 243. Shortest Word Distance 最短单词距离
[LeetCode] 245. Shortest Word Distance III 最短单词距离 III
[LeetCode] 244. Shortest Word Distance II 最短单词距离 II
原文：https://www.cnblogs.com/lightwindy/p/9736294.html