leetcode245- Shortest Word Distance III- medium

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.

 1.粗暴的就是if分开两种进行idx 的list的遍历处理。

2.直接扫描最开始的所有word的时候就记录下来。这道题目因为就是叫你求最小的距离嘛，那其实你一直跟踪着上一个出现过的单词位置就够了，因为这样它们总是追的最紧的，这样打擂台最后肯定能打出结果。这里有相同的word12输入的情况就是加一个小trick，p1 p2追踪的话，如果word1 == word2，那就把新出现的下标给p1，旧的下标给p2，这样两个的差值的确还是前后出现的这个数字的坐标差。

1.

class Solution {
    public int shortestWordDistance(String[] words, String word1, String word2) {
        
        List<Integer> idx1 = new ArrayList<>();
        List<Integer> idx2 = new ArrayList<>();

        for (int i = 0; i< words.length; i++) {
            if(word1.equals(words[i])) {
                idx1.add(i);
            } else if (word2.equals(words[i])){
                idx2.add(i);
            }
        }
        
        int result = Integer.MAX_VALUE;
        if (word1.equals(word2)) {
            for (int i = 1; i < idx1.size(); i++) {
                result = Math.min(result, idx1.get(i) - idx1.get(i - 1));
            }
            return result;
        }
        
        int p1 = 0, p2 = 0;
        while (p1 < idx1.size() && p2 < idx2.size()) {
            result = Math.min(result, Math.abs(idx1.get(p1) - idx2.get(p2)));
            if (idx1.get(p1) < idx2.get(p2)) {
                p1++;
            } else {
                p2++;
            }
        }
        return result;
    }
}

2.

class Solution {
    public int shortestWordDistance(String[] words, String word1, String word2) {
        
        int p1 = -1, p2 = -1;
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word1)) {
                p1 = word1.equals(word2) ? p2 : i;
            }
            if (words[i].equals(word2)) {
                p2 = i;
            }
            if (p1 != -1 && p2 != -1) {
                min = Math.min(min, Math.abs(p1 - p2));    
            }
        }
        return min;
    }
}