[CareerCup] 18.5 Shortest Distance between Two Words 两单词间的最短距离

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 = -, p2 = -, res = INT_MAX;
    for (int i = ; i < words.size(); ++i) {
        if (words[i] == word1) p1 = i;
        if (words[i] == word2) p2 = i;
        if (p1 != - && p2 != -) 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 = , j = , res = INT_MAX;
    for (int i = ; 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 = ; 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;
}

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