码迷,mamicode.com
首页 > 其他好文 > 详细

Substring with Concatenation of All Words

时间:2015-05-09 17:26:33      阅读:98      评论:0      收藏:0      [点我收藏+]

标签:

Substring with Concatenation of All Words

问题:

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in wordsexactly once and without any intervening characters.

思路:

  Hashtable + string API

我的代码:

技术分享
import java.util.Hashtable;

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> rst = new ArrayList<Integer>();
        if(s==null || s.length()==0 || words==null || words.length==0)  return rst;
        int wordLen = words[0].length();
        int count = words.length;
        Hashtable<String,Integer> ht = new Hashtable<String,Integer>();
        for(String word: words)
        {
            if(ht.containsKey(word)) ht.put(word, ht.get(word)+1);
            else ht.put(word,1);    
        }
        for(int i=0; i<=s.length()-count*wordLen; i++)
        {
            String sub = s.substring(i,i+count*wordLen);
            if(isValid(sub, ht, i, wordLen)) rst.add(i);
        }
        return rst;
    }
    public boolean isValid(String sub, Hashtable<String,Integer> ht, int start, int wordLen)
    {
        Hashtable<String,Integer> found = new Hashtable<String,Integer>();
        for(int i=0; i<=sub.length()-wordLen; i+=wordLen)
        {
            String key = sub.substring(i,i+wordLen);
            if(ht.containsKey(key))
            {
                if(found.containsKey(key))
                {
                    found.put(key,found.get(key)+1);
                    if(found.get(key) > ht.get(key))    return false;
                }
                else found.put(key,1);
            }
            else return false;
        }
        return true;
    }
}
View Code

 

学习之处:

  • String[] words可以包含重复的word,此处不适宜用Hashtable了,之前在微软的在线笔试中碰到了相应的问题,忘记考虑可以包含重复的元素了,丢了一些分数。
  • 既然包含重复的元素,怎么用hashtable进行存储呢,value解决方法有两个,一个是用List标记每一个重复的元素,另外一个使用count表明有几个重复的元素,第一种方式太过于复杂了,采用第二种方法,通过两个hashtable比较次数,判断是否由相应的word
  • 这道题是竟然暴力解法也能解决,注意命名规范
  • 改变不好的习惯 坏习惯+1

Substring with Concatenation of All Words

标签:

原文地址:http://www.cnblogs.com/sunshisonghit/p/4490536.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!