zoukankan      html  css  js  c++  java
  • LeetCode-Shortest Word Distance II

    This is a follow up of Shortest Word Distance. The only difference is now you are given the list of words and your method will be called repeatedly many times with different parameters. How would you optimize it?

    Design a class which receives a list of words in the constructor, and implements a method that takes two words word1 and word2 and 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.

    Solution:

    Use HashMap to store the indexs of each word.

     1 public class WordDistance {
     2     Map<String, List<Integer>> wordMap;
     3 
     4     public WordDistance(String[] words) {
     5         wordMap = new HashMap<String, List<Integer>>();
     6         for (int i = 0; i < words.length; i++) {
     7             String word = words[i];
     8             if (wordMap.containsKey(word)) {
     9                 wordMap.get(word).add(i);
    10             } else {
    11                 List<Integer> list = new ArrayList<Integer>();
    12                 list.add(i);
    13                 wordMap.put(word, list);
    14             }
    15         }
    16     }
    17 
    18     public int shortest(String word1, String word2) {
    19         // Get two index lists
    20         List<Integer> indList1 = wordMap.get(word1);
    21         List<Integer> indList2 = wordMap.get(word2);
    22         int p1 = 0, p2 = 0;
    23         int minDis = Integer.MAX_VALUE;
    24         while (true) {
    25             if (indList1.get(p1) < indList2.get(p2)) {
    26                 // Move p1.
    27                 while (p1 < indList1.size() && indList1.get(p1) < indList2.get(p2)) {
    28                     minDis = Math.min(minDis, Math.abs(indList1.get(p1) - indList2.get(p2)));
    29                     p1++;
    30                 }
    31                 if (p1 >= indList1.size())
    32                     break;
    33             } else {
    34                 // Move p2.
    35                 while (p2 < indList2.size() && indList2.get(p2) < indList1.get(p1)) {
    36                     minDis = Math.min(minDis, Math.abs(indList1.get(p1) - indList2.get(p2)));
    37                     p2++;
    38                 }
    39                 if (p2 >= indList2.size())
    40                     break;
    41             }
    42         }
    43         return minDis;
    44     }
    45 }
    46 
    47 
    48 // Your WordDistance object will be instantiated and called as such:
    49 // WordDistance wordDistance = new WordDistance(words);
    50 // wordDistance.shortest("word1", "word2");
    51 // wordDistance.shortest("anotherWord1", "anotherWord2");
  • 相关阅读:
    prepareStatement的用法和解释
    java socket报文通信(一) socket的建立
    java多线程小结
    Java_XML操作
    socket实例2
    socket实例1
    Socket小结
    从源码角度理解android动画Interpolator类的使用
    android使用属性动画代替补间动画
    OKHttp的简单使用
  • 原文地址:https://www.cnblogs.com/lishiblog/p/5798926.html
Copyright © 2011-2022 走看看