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.
Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.
题目标签：Array
题目给了我们一个words array， word1 和word2， 让我们找出word1 和word2 之间最短的距离。words array 里的word 可以重复出现。
基本思想是：当找到任何一个word 的时候，记录它的 index， 当两个word 都确定被找到的时候，用它们的index 相减来记录它们之间的距离
Java Solution:
Runtime beats 52.07%
完成日期：04/27/2017
关键词：Array
关键点：当找到任何一个word 时候，记录它的index；只有当两个word 都被找到的情况下，记录它们的距离
1 public class Solution 2 { 3 public int shortestDistance(String[] words, String word1, String word2) 4 { 5 int p1 = -1, p2 = -1, min = Integer.MAX_VALUE; 6 7 for(int i=0; i<words.length; i++) 8 { 9 if(words[i].equals(word1)) // if find word1, mark its position 10 p1 = i; 11 else if(words[i].equals(word2)) // if find word2, mark its position 12 p2 = i; 13 14 if(p1 != -1 && p2 != -1) // if find word1 and word2, save the smaller distance 15 min = Math.min(min, Math.abs(p1 - p2)); 16 } 17 18 return min; 19 } 20 }
参考资料：
https://discuss.leetcode.com/topic/20668/ac-java-clean-solution
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