[CareerCup] 18.5 Shortest Distance between Two Words 两单词间的最短距离-优快云博客

本文提供了解决LeetCode上短词距离问题的三种不同算法实现方案，包括一次调用的简单遍历方法，多次调用的预处理哈希映射方法，以及应对相同单词的情况。

18.5 You have a large text file containing words. Given any two words, find the shortest distance (in terms of number of words) between them in the file. If the operation will be repeated many times for the same file (but different pairs of words), can you optimize your solution?

LeetCode上的原题，请参见我之前的博客Shortest Word Distance，Shortest Word Distance II和 Shortest Word Distance III。

解法一：

// Call One Time
int shortest_dist(vector<string> words, string word1, string word2) {
    int p1 = -1, p2 = -1, res = INT_MAX;
    for (int i = 0; i < words.size(); ++i) {
        if (words[i] == word1) p1 = i;
        if (words[i] == word2) p2 = i;
        if (p1 != -1 && p2 != -1) res = min(res, abs(p1 - p2));
    }
    return res;
}

解法二：

// Call Many Times
int shortest_dist(vector<string> words, string word1, string word2) {
    unordered_map<string, vector<int>> m;
    int i = 0, j = 0, res = INT_MAX;
    for (int i = 0; i < words.size(); ++i) {
        m[words[i]].push_back(i);
    }
    while (i < m[word1].size() && j < m[word2].size()) {
        res = min(res, abs(m[word1][i] - m[word2][j]));
        m[word1][i] < m[word2][j] ? ++i : ++j;
    }
    return res;
}

解法三：

// word1, word2 may be same
int shortest_dist(vector<string> words, string word1, string word2) {
    int p1 = words.size(), p2 = -words.size(), res = INT_MAX;
    for (int i = 0; i < words.size(); ++i) {
        if (words[i] == word1) p1 = word1 == word2 ? p2 : i;
        if (words[i] == word2) p2 = i;
        res = min(res, abs(p1 - p2));
    }
    return res;
}