Shortest Word Distance III

原创已于 2022-02-25 15:18:55 修改 · 374 阅读

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于 2016-09-10 04:45:32 首次发布

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本文介绍了一种算法，用于寻找列表中两个指定单词之间的最短距离，特别关注于处理两个单词相同的情况。通过迭代列表并跟踪最近出现的单词位置来实现。

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.

思路：这题跟之前Short Word Distance的区别在于，index1 index2都可以更新，但是如果两者相等的时候，就不要update distance；只有不相等的时候update就可以了。

class Solution {
    public int shortestWordDistance(String[] wordsDict, String word1, String word2) {
         int index1 = -1;
        int index2 = -1;
        int minDistance = Integer.MAX_VALUE;
        for(int i = 0; i < wordsDict.length; i++) {
            String word = wordsDict[i];
            if(word.equals(word1)) {
                index1 = i;
                if(index2 != -1 && index1 != index2) {
                    minDistance = Math.min(minDistance, index1 - index2);
                }
            }
            if(word.equals(word2)) {
                index2 = i;
                if(index1 != -1 && index1 != index2) {
                    minDistance = Math.min(minDistance, index2 - index1);
                }
            }
        }
        return minDistance;
    }
}