This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.

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

word1 and word2 may be the same and they represent two individual words in the list.

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

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

Note:
You may assume word1 and word2 are both in the list.

這道題還是讓我們求最短單詞距離，有了之前兩道題Shortest Word Distance II和

解法一：

class Solution {
 
public:
    int shortestWordDistance(vector<string>& words, string word1, string word2) {
        int p1 = -1, p2 = -1, res = INT_MAX;
        for (int i = 0; i < words.size(); ++i) {
            int t = p1;
            if (words[i] == word1) p1 = i;
            if (words[i] == word2) p2 = i;
            if (p1 != -1 && p2 != -1) {
                if (word1 == word2 && t != -1 && t != p1) {
                    res = min(res, abs(t - p1));
                } else if (p1 != p2) {
                    res = min(res, abs(p1 - p2));
                }
            }
        }
        return res;
    }
};

上述程式碼其實可以優化一下，我們並不需要變數t來記錄上一個位置，我們將p1初始化為陣列長度，p2初始化為陣列長度的相反數，然後當word1和word2相等的情況，我們用p1來儲存p2的結果，p2賦為當前的位置i，這樣我們就可以更新結果了，如果word1和word2不相等，則還跟原來的做法一樣，這種思路真是挺巧妙的，參見程式碼如下：

解法二：

class Solution {
public:
    int shortestWordDistance(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;
    }
};

我們再來看一種更進一步優化的方法，只用一個變數idx，這個idx的作用就相當於記錄上一次的位置，當前idx不等-1時，說明當前i和idx不同，然後我們在word1和word2相同或者words[i]和words[idx]相同的情況下更新結果，最後別忘了將idx賦為i，參見程式碼如下；

解法三：

class Solution {
public:
    int shortestWordDistance(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 && (word1 == word2 || words[i] != words[idx])) {
                    res = min(res, i - idx);
                }
                idx = i;
            }
        }
        return res;
    }
};

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