[LeetCode] Shortest Word Distance 最短单词距离

 

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.

 

这道题让咱们给了咱们一个单词数组，又给定了两个单词，让咱们求这两个单词之间的最小距离，限定了两个单词不一样，并且都在数组中。我最早想到的方法比较笨，我首先想的是要用哈希表来作，创建每一个单词和其全部出现位置数组的映射，可是后来想一想，反正创建映射也要遍历一遍数组，咱们还不如直接遍历一遍数组，直接把两个给定单词全部出现的位置分别存到两个数组里，而后咱们在对两个数组进行两两比较更新结果，参见代码以下：

 

解法一：

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

 

上面的那种方法并不高效，咱们其实须要遍历一次数组就能够了，咱们用两个变量p1,p2初始化为-1，而后咱们遍历数组，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，若是此时p1, p2都不为-1了，那么咱们更新结果，参见代码以下：

 

解法二：

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;
    }
};

 

下面这种方法只用一个辅助变量idx，初始化为-1，而后遍历数组，若是遇到等于两个单词中的任意一个的单词，咱们在看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一样，咱们更新结果，参见代码以下：

 

解法三：

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[idx] != words[i]) {
                    res = min(res, i - idx);
                }
                idx = i;
            }
        }
        return res;
    }
};

 

相似题目：

Shortest Word Distance II

Shortest Word Distance III 

 

参考资料：

https://leetcode.com/discuss/50234/ac-java-clean-solution

https://leetcode.com/discuss/61820/java-only-need-to-keep-one-index

 

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