动态规划、递归：word-break II

动态规划、递归的典例

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”].

让你把字符串拆解，每个拆解部分都是右侧Set里的元素

析：根据上一题可以求出所有的下标j,使得从(j,末尾)满足是集合中的元素。对于字符串s，如果s整体在集合中，加入List，之后在s中找断开的位置，将满足的字符串加入到List,
本题用到： 动态规划与递归

package com.yixin.exam;import java.util.*;public class Main {    public static void main(String[] args){        Scanner sc= new Scanner(System.in);        Main main = new Main();        String s="catsanddog";           String[] ss ={"cat", "cats", "and", "sand", "dog"};        Set<String> dict = new HashSet<>();        for(String t:ss)            dict.add(t);        System.out.println(main.wordBreak(s,dict));    }    public ArrayList<String> wordBreak(String s, Set<String> dict) {        List<Integer> indexList = wordBreak2(s,dict);        ArrayList<String> res = wordBreak1(s,dict,0,indexList);        return res;    }    // 注：比如求出的indexList为7，4，2，0；    // firstIndex为当前需要分割的字串的首字母的下标    public ArrayList<String> wordBreak1(String s, Set<String> dict,int firstIndex,List<Integer> indexList) {        ArrayList<String> res = new ArrayList<>();        String subStr = s.substring(firstIndex);        if(dict.contains(subStr))            res.add(subStr);        //        int pos = indexList.indexOf(firstIndex);        for(int j=pos-1;j>=0;j--){            // laterIndex表示下一个可以分割的位置，比如0之后是2，2之后是4            int laterIndex = indexList.get(j);            subStr =s.substring(firstIndex,laterIndex);            if(dict.contains(subStr)){                ArrayList<String> innerRes = wordBreak1(s,dict,laterIndex,indexList);                for(String t:innerRes){                    res.add(subStr+" "+t);                }            }        }        return res;    }    //  求出能够分割的点使点的右侧在集合中，参见上一题    public List<Integer> wordBreak2(String s, Set<String> dict) {        int n = s.length();        List<Integer> indexList = new ArrayList<>();        for(int i=n-1;i>=0;i--){            if(dict.contains(s.substring(i)))                indexList.add(i);            else                for(Integer t:indexList){                    if(dict.contains(s.substring(i,t))){                        indexList.add(i);                        break;                    }                }        }        return indexList;    }}