243. Shortest Word Distance (Easy)

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

For example,

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

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

Note:

You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.

Solution 1: Array 15ms

这道题让我们给了我们一个单词数组，又给定了两个单词，让我们求这两个单词之间的最小距离，限定了两个单词不同，而且都在数组中。我最先想到的方法比较笨，我首先想的是要用哈希表来做，建立每个单词和其所有出现位置数组的映射，但是后来想想，反正建立映射也要遍历一遍数组，我们还不如直接遍历一遍数组，直接把两个给定单词所有出现的位置分别存到两个数组里，然后我们在对两个数组进行两两比较更新结果，参见代码如下：

class Solution {
public:
    int shortestDistance(vector<string>& words, string word1, string word2) {
        vector<int> idx1, idx2;
        for (int i = 0; i < words.size(); ++i) {
            if (words[i] == word1) idx1.push_back(i);
            else if (words[i] == word2) idx2.push_back(i);
        }
        int dist = INT_MAX;
        for (int i = 0; i < idx1.size(); ++i) {
            for (int j = 0; j < idx2.size(); ++j) {
                dist = min(dist, abs(idx1[i]-idx2[j]));
            }
        }
        return dist;
    }
};

Solution 2: 15ms

上面的那种方法并不高效，我们其实需要遍历一次数组就可以了，我们用两个变量p1,p2初始化为-1，然后我们遍历数组，遇到单词1，就将其位置存在p1里，若遇到单词2，就将其位置存在p2里，如果此时p1, p2都不为-1了，那么我们更新结果，参见代码如下：

class Solution {
public:
    int shortestDistance(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;
            else if (words[i] == word2) p2 = i;
            if (p1 != -1 && p2 != -1) res = min(res, abs(p1-p2));
        }
        return res;
    }
};

Solution 3: 16ms

我们还可以进一步优化上面的算法，只用一个辅助变量idx，初始化为-1，然后遍历数组，如果遇到等于两个单词中的任意一个的单词，我们在看idx是否为-1，若不为-1，且指向的单词和当前遍历到的单词不同，我们更新结果，参见代码如下：

class Solution {
public:
    int shortestDistance(vector<string>& words, string word1, string word2) {
        int idx = -1, res = INT_MAX;
        for (int i = 0; i < words.size(); ++i) {
            if (words[i] == word1 || words[i] == word2) {
                if (idx != -1 && words[i] != words[idx]) {
                    res = min(res, i-idx);
                }
                idx = i;
            }
        }
        return res;
    }
};