• 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"].

    class Solution {
    public:
        vector<string> wordBreak(string s, unordered_set<string> &dict) 
        {
            vector<string> result;
            if(check(s,dict)==false) return result;
            
            int v[s.length()+1];v[0]=0;
            generate(result,s,dict,v,1);
            return result;
        }
        bool check(string s, unordered_set<string> &dict) 
        {
            bool able[s.length()+1];
            able[0]=true;
            for(int i=1;i<=s.length();i++)
            {
                able[i]=false;
                for(int j=0;j<i;j++)
                if(able[j] && dict.find(s.substr(j,i-j))!=dict.end())
                {
                    able[i]=true;
                    continue;
                }
            }
            
            return able[s.length()];
        }
        void generate(vector<string>& result,const string& s,const unordered_set<string> &dict,int* v,int vdep)
        {
            if(v[vdep-1]==s.length())
            {
                string snew=s.substr(0,v[1]);
                for(int i=2;i<vdep;i++)
                {
                    snew=snew+" "+s.substr(v[i-1],v[i]-v[i-1]);
                }
                result.push_back(snew);
                return;
            }
            for(int i=1;i<=s.length()-v[vdep-1];i++)
                if(dict.find(s.substr(v[vdep-1],i))!=dict.end())
                {
                    v[vdep]=v[vdep-1]+i;
                    generate(result,s,dict,v,vdep+1);
                }
        }
    };
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  • 原文地址:https://www.cnblogs.com/erictanghu/p/3759708.html
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