243. Shortest Word Distance

最新推荐文章于 2020-10-04 23:11:00 发布

原创最新推荐文章于 2020-10-04 23:11:00 发布 · 178 阅读

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本文介绍了一种高效算法，用于计算给定单词列表中两个特定单词之间的最短距离。通过跟踪最近遇到的两个目标单词的位置，该算法能够在一次遍历内解决问题。

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.

一开始想复杂了总想直接找到最相近的两个索引

    public int shortestDistance(String[] words, String word1, String word2) {
        List<Integer> list = new ArrayList<>();
        TreeSet<Integer> set = new TreeSet<>();
        for (int i=0; i<words.length; i++) {
            if (words[i].equals(word1)) {
                list.add(i);
            } else if (words[i].equals(word2)) {
                set.add(i);
            }
        }
        
        int min = Integer.MAX_VALUE;
        for (Integer n : list) {
            Integer small = set.floor(n);
            if (small != null) min = Math.min(min, n-small);
            Integer big = set.ceiling(n);
            if (big != null) min = Math.min(min, big-n);
        }
        return min;
    }

实际上不需要这样只要找出所有word的索引遍历求差值就可以了

public int shortestDistance(String[] words, String word1, String word2) {
    int i1 = -1, i2 = -1;
    int minDistance = words.length;
    int currentDistance;
    for (int i = 0; i < words.length; i++) {
        if (words[i].equals(word1)) {
            i1 = i;
        } else if (words[i].equals(word2)) {
            i2 = i;
        }

        if (i1 != -1 && i2 != -1) {
            minDistance = Math.min(minDistance, Math.abs(i1 - i2));
        }
    }
    return minDistance;
}