[leetcode] 244. Shortest Word Distance II 解题报告

题目链接: https://leetcode.com/problems/shortest-word-distance-ii/
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 calledrepeatedly 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.

思路: 每一个单词可能出现多次, 然后将其位置存在以其字符串为key的hash表中, 然后遍历两个字符串的各个位置, 找出最短距离.在找两个字符串最短间距的时候可以naive的方式是枚举所有位置,可以优化的地方是在比较的时候让低位的移动.这样时间复杂度可以由O(m*n)降到O(m+n).

代码如下:

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

int shortest(string word1, string word2) {
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;
}
private:
unordered_map<string, vector<int>> hash;
};

// Your WordDistance object will be instantiated and called as such:
// WordDistance wordDistance(words);
// wordDistance.shortest("word1", "word2");
// wordDistance.shortest("anotherWord1", "anotherWord2");

参考:https://leetcode.com/discuss/51698/9-line-o-n-c-solution