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;
}
};