[LeetCode] 243. Shortest Word Distance 最短單詞距離

 

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

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

Input: word1 = “coding”, word2 = “practice”
Output: 3

Input: word1 = "makes", word2 = "coding"
Output: 1

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

 

這道題給了我們一個單詞數組，又給定了兩個單詞，讓求這兩個單詞之間的最小距離，限定了兩個單詞不同，而且都在數組中。博主最先想到的方法比較笨，是要用 HashMap 來做，建立每個單詞和其所有出現位置數組的映射，但是后來想想，反正建立映射也要遍歷一遍數組，還不如直接遍歷一遍數組，直接把兩個給定單詞所有出現的位置分別存到兩個數組里，然后再對兩個數組進行兩兩比較更新結果，參見代碼如下：

 

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;
    }
};

 

上面的那種方法並不高效，其實需要遍歷一次數組就可以了，用兩個變量 p1,p2 初始化為 -1，然后遍歷數組，遇到單詞1，就將其位置存在 p1 里，若遇到單詞2，就將其位置存在 p2 里，如果此時 p1, p2 都不為 -1 了，那么更新結果，參見代碼如下：

 

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;
    }
};

 

下面這種方法只用一個輔助變量 idx，初始化為 -1，然后遍歷數組，如果遇到等於兩個單詞中的任意一個的單詞，再看 idx 是否為 -1，若不為 -1，且指向的單詞和當前遍歷到的單詞不同，更新結果，參見代碼如下：

 

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;
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/243

 

類似題目：

Shortest Word Distance II

Shortest Word Distance III 

 

參考資料：

https://leetcode.com/problems/shortest-word-distance/

https://leetcode.com/problems/shortest-word-distance/discuss/66931/AC-Java-clean-solution

https://leetcode.com/problems/shortest-word-distance/discuss/66939/Java%3A-only-need-to-keep-one-index

 

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