[leetcode] 245. Shortest Word Distance III 解题报告

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本文探讨了如何解决LeetCode中的'短语距离III'问题，该问题涉及查找同一列表中两个单词之间的最短距离。文章通过使用哈希表来存储每个单词的位置，进而比较不同或相同单词的不同位置，以找到最短距离。特别地，当两个给定的单词相同时，文章强调了如何正确处理索引差异。

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

This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same asword2.

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

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

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

Note:
You may assume word1 and word2 are both in the list.

思路: 和上题一样, 用hash表将每一个单词的位置保存起来, 因为一个单词可能在字典中出现多次, 因此hash表对应一个单词的是一个位置集合.最后将两个单词的不同位置做比较即可.需要注意的是如果给定的两个单词是相同的,则要判断索引是不同的.

代码如下;

class Solution {
public:
    int shortestWordDistance(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, ans = INT_MAX;
        while(i < hash[word1].size() && j < hash[word2].size())
        {
            if(hash[word1][i] != hash[word2][j])
                ans = min(ans, abs(hash[word1][i] - hash[word2][j]));
            hash[word1][i]<hash[word2][j]?i++:j++;
        }
        return ans;
    }
};