[LeetCode]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"].

Analysis:

For the "Return all " problems, usually DFS or BFS will work well.

In this problem, a DFS with a simple cutting edge condition will pass all the test cases.

The idea is to use mp[][] array to record which substring is in dict. and a array to record the insert index

for those substrings, we use DFS. 

c++

class Solution {public:    void dfsWordBreak(string &s,                  int st,                  vector<int>&sp,                  unordered_set<string> &dict,                  vector<string> &result,                  vector<vector<bool>> &mp){    if(st>=s.size()){        string str = s;        for(int i=0;i<sp.size()-1;i++){            str.insert(sp[i]+i," ");        }        result.push_back(str);    }else{        for(int j=0;j<mp[st].size();j++){            if(mp[st][j]==true){                sp.push_back(j+1);                dfsWordBreak(s,j+1,sp,dict,result,mp);                sp.pop_back();            }        }    }}vector<string> wordBreak(string s, unordered_set<string> &dict) {    vector<string> result;    vector<vector<bool>> mp(s.size(),vector<bool>(s.size(),false));    for(int i=0;i<s.size();i++){        for(int j=i;j<s.size();j++){            if(dict.find(s.substr(i,j-i+1))!=dict.end())                mp[i][j] = true;        }    }    bool flag = false;    for(int i=0;i<s.size();i++){        if(mp[i][s.size()-1])            {flag = true;            break;}    }    if(!flag) return result;    vector<int>sp;    dfsWordBreak(s,0,sp,dict,result,mp);    return result;}};

Java

public class Solution {    List<String> result;List<Integer> solu;boolean [][] mp;public List<String> wordBreak(String s, Set<String> dict) {        result = new ArrayList<>();        solu = new ArrayList<>();        mp = new boolean[s.length()][s.length()];        for(int i=0;i<s.length();i++){        for(int j=i;j<s.length();j++){        if(dict.contains(s.substring(i, j+1)))        mp[i][j] = true;        }        }        boolean flag = false;        for(int i=0;i<s.length();i++){        if(mp[i][s.length()-1]){        flag = true;        break;        }        }        if(!flag) return result;        dfs(s, 0, dict);        return result;    }public void dfs(String s, int st, Set<String> dict ){if(st>=s.length()){StringBuffer str = new StringBuffer(s);for(int i=0;i<solu.size()-1;i++){str.insert(solu.get(i)+i, " ");}result.add(str.toString());}else {for(int j=0;j<mp[st].length;j++){if(mp[st][j]==true){solu.add(j+1);dfs(s, j+1, dict);solu.remove(solu.size()-1);}}}}}