动态规划——word-break&&word-breakii

题目描述

Given a string s and a dictionary of wordsdict, determine ifs can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s ="leetcode",
dict =["leet", "code"].

Return true because"leetcode"can be segmented as"leet code".

思路：dp[i]  表示源串的前i个字符可以满足分割，那么 dp[ j ] 满足分割的条件是存在i(i<j) 使得 dp [i] && substring(i,j)在字典里

(dp[i]和substring(i,j)中的i的意义不同,dp[i]表示s的前i个字符在字典里，substring的i表示字符串中的下标i指向第i+1个字符)。

这里需要注意：dp[i]的下标i表示s的第i个字符，而s的第i个字符在s的下标为i-1!!!

import java.util.*;
public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        if(s == null||dict == null)
            return false;
        //dp[i]存放s前i个字符是否满足条件；
        boolean[] dp=new boolean[s.length()+1];
        dp[0]=true;
        for(int i=1;i<=s.length();i++)
            {
            if(dp[i-1] == true)//前i-1个字符返回true;
                {
                  //s中第i个字符下标为i-1;s.substring(i-1,j);
                  for(int j=i;j<=s.length();j++)
                      {
                      if(dict.contains(s.substring(i-1,j)))
                          {
                          dp[j]=true;
                          if(j == s.length())
                              {
                              return true;
                          }
                      }
                  }
                }
            }
        return dp[s.length()];
     }
}

or:

import java.util.*;
public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        if(s == null||s.length() == 0||dict == null||dict.size() == 0){
            return false;
        }
        boolean[] dp=new boolean[s.length()+1];
        //dp[i]表示s长度为i时是否在dict中
        dp[0]=true;
        for(int i=1;i<=s.length();i++){
            for(int j=0;j<i;j++){
                if(dp[j] == true&&dict.contains(s.substring(j,i))){
                    dp[i]=true;
                }
            }
        }
        return dp[s.length()];
    }
}

题目描述:word-breakii

Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.

Return all such possible sentences.

For example, given
s ="catsanddog",
dict =["cat", "cats", "and", "sand", "dog"].

A solution is["cats and dog", "cat sand dog"].

标准的DFS套路题。

话不投机半句多！

import java.util.*;
public class Solution {
    public ArrayList<String> wordBreak(String s, Set<String> dict) {
         ArrayList<String> array=new  ArrayList<String>();
        if(s == null||dict == null||s.length() == 0||dict.size() == 0)
            return array;
        dfs(s,s.length(),"",array,dict);
        return array;
    }
    public void dfs(String s,int size,String result,ArrayList<String> array,Set<String> dict)
        {
        if(size == 0)
            {
            result=result.trim();
            array.add(result);
            return;
        }
        for(int i=size-1;i>=0;i--)
            {
            if(dict.contains(s.substring(i,size)))
                {
                dfs(s,i,s.substring(i,size)+" "+result,array,dict);
            }
        }
    }
}