[leetcode 244] Shortest Word Distance II------------字典中两个字符串的最短距离

Question:

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.

分析：
这道题的意思是，实现一个类，和给定两个字符串，字典中某个字符串可能出现多次，求这两个给定字符串的最短距离。

实现字典类的时候，用无序容器map实现，并存没给字符串所在的所有位置，这样给定两个字符串的时候就可以知道这两个字符串在字典中的所有位置。

代码如下（转网上大神）：

<span style="font-size:14px;">class WordDistance {
public:
WordDistance(vector<string> words) {
int n = words.size();
for (int i = 0; i < n; i++)
wordInd[words[i]].push_back(i);
}

int shortest(string word1, string word2) {
vector<int> indexes1 = wordInd[word1];
vector<int> indexes2 = wordInd[word2];
int m = indexes1.size(), n = indexes2.size();
int i = 0, j = 0, dist = INT_MAX;
while (i < m && j < n) {
dist = min(dist, abs(indexes1[i] - indexes2[j]));
if (indexes1[i] < indexes2[j]) i++;
else j++;
}
return dist;
}
private:
unordered_map<string, vector<int> > wordInd;
};
</span>