[Leetcode] 243. Shortest Word Distance 解题报告

题目：

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：

注意到word1和word2有可能在字典里面出现多次，我们可以用哈希表保存一个单词的多个位置，然后枚举最小距离。假设word1在words中出现了m次，word2在words中出现了n次，则枚举需要O(m*n)的时间复杂度，可是由于位置索引都是单调递增的，所以我们可以采用一个小技巧将时间复杂度降低到O(m+n)。这里卖个关子，大家看看下面的代码实现^_^。

代码：

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, ret = INT_MAX;
        while (i < hash[word1].size() && j < hash[word2].size()) {
            ret = min(ret, abs(hash[word1][i] - hash[word2][j]));
            hash[word1][i] < hash[word2][j] ? ++i : ++j;
        }
        return ret;
    }
};