140. Word Break II(dp)

来源：互联网发布：c语言编写可视化界面编辑：程序博客网时间：2024/04/27 18:08

题目：

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"].

同Word Break，要将具体分割表示出来。

思路：

采用dp求出map[i][j],表示i到j在词典中。

采用深搜，求出每一种分割。

代码：

public class Solution {    public List<String> wordBreak(String s, Set<String> wordDict) {        List<String> result = new ArrayList<String>();        List<String> path = new ArrayList<String>();        boolean[] dp = new boolean[s.length()+1];        boolean[][] map = new boolean[s.length()+1][s.length()+1];        dp[0] = true;                for(int i=0;i<s.length();i++)        {        if(dp[i])        {        for(int len = 1;len<s.length()-i+1;len++)        {        if(wordDict.contains(s.substring(i, i+len)))        {        dp[i+len]= true;        map[i][i+len]=true;        }        }        }        }        if(dp[s.length()]){            dfs(map,result,path,0,s);        }                return result;            }        private void dfs(boolean[][] map,List<String> result,List<String> path,int start,String s)    {    if(start == s.length())    {    String temp="";    for(String str:path)    {    temp=temp+str+" ";    }    result.add(temp.substring(0, temp.length()-1));    }        for(int i=0;i<s.length()+1;i++)    {    if(map[start][i])    {    path.add(s.substring(start, i));    dfs(map,result,path,i,s);    path.remove(path.size()-1);    }    }    }}

0 0