Leetcode - Shortest Word Distance III

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本文探讨了如何计算单词列表中两个单词之间的最短距离，包括同一单词的情况。提供了两种方法来解决这个问题，并通过实例演示了算法的应用。

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.


public class Solution {
    // Method 1
    public int shortestWordDistance1(String[] words, String word1, String word2) {
        if (words == null) return -1;
        if (word1.equals(word2)) return shortestWordDistanceCaseSame(words, word1);

        int idx1 = -1, idx2 = -1;
        int diff = words.length;
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word1)) {
                idx1 = i;
                if (idx2 != -1) {
                    diff = Math.min(diff, idx1 - idx2);
                }
            } else if (words[i].equals(word2)) {
                idx2 = i;
                if (idx1 != -1) {
                    diff = Math.min(diff, idx2 - idx1);
                }
            }
        }
        return diff;
    }
    public int shortestWordDistanceCaseSame(String[] words, String word) {
        int prev = -1, curr = -1;
        int diff = words.length;
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word)) {
                curr = i;
                if (prev != -1)
                    diff = Math.min(curr - prev, diff);
                prev = curr;
            }
        }
        return diff;
    }

    // Method 2: all in one method
    public int shortestWordDistance(String[] words, String word1, String word2) {
        if (words == null) return -1;
        int idx1 = -1, idx2 = -1;
        int diff = words.length;
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word1)) {
                if (word1.equals(word2)) {
                    if (idx1 != -1) 
                        diff = Math.min(diff, i - idx1);
                    idx1 = i;
                } else {
                    idx1 = i;
                    if (idx2 != -1) {
                        diff = Math.min(diff, idx1 - idx2);
                    }
                }
            } else if (words[i].equals(word2)) {
                idx2 = i;
                if (idx1 != -1) {
                    diff = Math.min(diff, idx2 - idx1);
                }
            }
        }
        return diff;
    }
}