Shortest Word Distance III

来源:互联网 发布:轩辕剑纹身进阶数据 编辑:程序博客网 时间:2024/06/08 06:58

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。

public class Solution {    public int shortestWordDistance(String[] words, String word1, String word2) {        if(words == null || words.length == 0 || word1 == null || word2 == null) return 0;        int index1 = -1; int index2 = -1;        int mindis = Integer.MAX_VALUE;        for(int i=0; i<words.length; i++){            if(words[i].equals(word1)){                index1 = i;                if(index2 != -1){                    mindis = index1 != index2 ? Math.min(mindis, index1 - index2) : mindis;                }            }            if(words[i].equals(word2)){                index2 = i;                if(index1 != -1){                    mindis = index2 != index1 ? Math.min(mindis, index2 - index1) : mindis;                }            }        }        return mindis;    }}


0 0
原创粉丝点击