Leetcode: word-break

题目一：word-break

Given a string s and a dictionary of words dict, determine if s 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：

1.动态规划，利用dp保存状态，将字符串切割成小段

2.双层循环，判断当前是否能被有效分隔

代码：

import java.util.*;
public class Solution {
    public boolean wordBreak(String s, Set<String> dict) {
        boolean[] dp = new boolean[s.length()+1];
        dp[0] = true;
        for(int i=1; i<=s.length(); i++)
        {
            for(int j=0; j<i; j++)
            {
                if(dp[j] && dict.contains(s.substring(j,i)))
                {
                    dp[i] = true;
                }
            }
        }
        return dp[s.length()];
    }
}

题目二：word-break-ii

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"].

思路：

1.将字符串递归分成左右两个部分s1，s2.若s1在字典里，则递归计算s2

2.当s2无法分割时，返回空ArrayList

3.当返回的ArrayList不为空是，将结果保存到ret

4.用map来记录递归状态，用于递归结束条件

代码：

import java.util.*;
public class Solution {
    Map<String, ArrayList<String>> map = new HashMap<String, ArrayList<String>>();
    public ArrayList<String> wordBreak(String s, Set<String> dict) {   
        if(map.containsKey(s))
            return map.get(s);
        ArrayList<String> ret = new ArrayList<String>();
        if(dict.contains(s))
            ret.add(s);
        for(int i=0; i<s.length(); i++)
        {
            String cur = s.substring(i);
            if(dict.contains(cur))
            {
                ArrayList<String> strs = wordBreak(s.substring(0,i), dict);
                if(strs.size() > 0)
                {
                    for(String str : strs)
                    {
                        ret.add(str + " " + cur);
                    }
                }
            }
        }
        map.put(s,ret);
        return ret; 
    }
}