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

题目：

This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.

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：

和上面两道题目依然非常类似，唯一区别就是：在枚举之前先判断一下两者的索引是否一样，一旦一样，就说明word1和word2是相同的，此时只要让word1的索引加1或者word2的索引加1就可以了。时间复杂度和空间复杂度与上面两道题目完全一样。

代码：

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