LeetCode-Word Break II

题目：https://oj.leetcode.com/problems/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"]

.
源码：Java版本

算法分析：动态规划。时间复杂度O(n^2)，空间复杂度O(n^2)

public class Solution {
    public List<String> wordBreak(String s, Set<String> dict) {
        boolean[] f=new boolean[s.length()+1];
        //b[i][j] 为true，表示s[j, i) 是一个合法单词，可以从j 处切开
        boolean[][] b=new boolean[s.length()+1][s.length()];
        f[0]=true;
        
        for(int i=1;i<=s.length();i++) {
            for(int j=i-1;j>=0;j--) {
                if(f[j]&&dict.contains(s.substring(j,i))) {
                    f[i]=true;
                    b[i][j]=true;
                }
            }
        }
        List<String> results=new ArrayList<String>();
        Stack<String> path=new Stack<String>();
        genPath(s,b,s.length(),path,results);    
        return results;
    }
    
    //DFS 遍历树，生成路径
    private void genPath(String s,boolean[][] b,int current,Stack<String> path,List<String> results){
        if(current==0) {
            Stack<String> temp=(Stack<String>)(path.clone());
		    String str="";
		    if(!temp.isEmpty()) {
			    str+=temp.pop();
		    }
		    while(!temp.isEmpty()) {
			    str+=" "+temp.pop();
		    }
	        results.add(str);
            return;    
        }
        
        for(int i=0;i<current;i++) {
            if(b[current][i]) {
                path.add(s.substring(i,current));
                genPath(s,b,i,path,results);
                path.pop();
            }
        }
    }  
}