245. Shortest Word Distance III

最新推荐文章于 2020-07-22 09:38:11 发布

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最新推荐文章于 2020-07-22 09:38:11 发布

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分类专栏： LeetCode Array 文章标签： LeetCode Array Shortest Word Distan

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/60326491

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本文介绍了一种使用双指针法解决最短单词距离问题的方法，特别针对word1和word2可能相同的情况。通过遍历字符串数组并更新两个指针的位置来计算最短距离。

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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.

用双指针，初始化为-1，找到word1，赋值w1，找到word2，赋值w2，如果两个相等，在w1的处理函数里进行距离比较，如果不等，在两个word都判断完之后比较。代码如下：

public class Solution {
    public int shortestWordDistance(String[] words, String word1, String word2) {
        int w1 = -1, w2 = -1, max = Integer.MAX_VALUE;
        for (int i = 0; i < words.length; i ++) {
            if (word1.equals(words[i])) {
                if (w1 != -1 && word1.equals(word2)) {
                    max = Math.min(max, Math.abs(i - w1));
                }
                w1 = i;
            } else if (word2.equals(words[i])) {
                w2 = i;
            }
            if (w1 != -1 && w2 != -1)
                    max = Math.min(max, Math.abs(w1 - w2));
        }
        return max;
    }
}