lintcode:单词切分

单词切分

给出一个字符串s和一个词典，判断字符串s是否可以被空格切分成一个或多个出现在字典中的单词。

样例

s = "lintcode"

dict = ["lint","code"]

返回 true 因为"lintcode"可以被空格切分成"lint code"

解题

DFS

import java.util.Iterator;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeSet;
// write your code here
public class Main{
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        Main m = new Main();
        while(in.hasNext()){
            String s = in.nextLine();
            String[] str = in.nextLine().split(" ");
            Set<String> dict = new TreeSet<String>();
            for(int i = 0;i<str.length;i++)
                dict.add(str[i]);
            int start = 0;
            boolean flag = m.wordBreak(s, dict, start);
            System.out.println(flag);
        }
    }
    public boolean wordBreak(String s,Set<String> dict,int start){

　　　　　　if((s==null ||s.length() ==0) && (dict == null || dict.size()==0))
　　　　　　　　return true;

        Iterator it = dict.iterator();
        if(start == s.length())
            return true;
        while(it.hasNext()){
            String t = (String)it.next();
            int end = start + t.length();
            if(end > s.length())
                continue;
            if(s.substring(start,end).equals( t )){
                if(wordBreak(s,dict,end)){
                    return true;
                }
            }
        }
        return false;
    }
}

95%数据运行超时

动态规划求解

定义数组dp dp[i] =true表示 字符串 s 子串0 - (i-1)在字典中存在

当dp[s.length()] == true 时候表示可以由字典内的单词组成s

public class Solution {
    /**
     * @param s: A string s
     * @param dict: A dictionary of words dict
     */
    public boolean wordBreak(String s, Set<String> dict) {
        // write your code here
        if((s==null ||s.length() ==0) && (dict == null || dict.size()==0))
            return true;
        return wordBreak(s,dict,0);
    }
     public boolean wordBreak(String s,Set<String> dict,int start){
        boolean dp[] = new boolean[s.length() + 1];
        dp[0] = true;//初始值
        for(int i = 0;i<s.length();i++){
            if(!dp[i])
                continue;
            for(String t:dict){
                int len = t.length();
                int end = i+ len;
                if(end > s.length())
                    continue;
                if(s.substring(i,end).equals(t)){
                    dp[end] = true;
                }
            }
        }
        return dp[s.length()];
    }

}