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"].

[解题思路]

DFS + backtracking

在大数据："aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]通不过。

C++实现代码：

#include<iostream>
#include<unordered_set>
#include<string>
#include<cstring>
using namespace std;

class Solution
{
public:
    bool wordBreak(string s, unordered_set<string> &dict)
    {
        int len=s.length();
        if(len==0)
            return true;
        bool dp[len+1];
        memset(dp,0,sizeof(dp));
        int i,j;
        for(i=1; i<=len; i++)
        {
            if(dp[i]==false&&dict.count(s.substr(0,i))>0)
                dp[i]=true;
            if(i==len&&dp[i]==true)
                break;
            if(dp[i]==true)
            {
                for(j=i+1; j<=len; j++)
                {
                    if(dp[j]==false&&dict.count(s.substr(i,j-i))>0)
                        dp[j]=true;
                    if(j==len&&dp[j]==true)
                        break;
                }
            }
        }
        if(i==len||j==len)
            return dp[i]||dp[j];
        return false;
    }
};

int main()
{
    unordered_set<string> dict= {"cat", "cats", "and", "sand", "dog"};
    Solution s;
    string ss = "catsanddog";
    cout<<s.wordBreak(ss,dict)<<endl;
}

 2.DP+DFS

使用Word Break对字符串进行处理，处理结束后使用DFS进行DFS。

#include<iostream>
#include<unordered_set>
#include<string>
#include<cstring>
#include<vector>
using namespace std;

class Solution
{
public:
    vector<string> ss;
    vector<string> wordBreak(string s, unordered_set<string> &dict)
    {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        int n=s.size();
        ss.clear();
        bool dp[n];
        memset(dp,false,sizeof(dp));
        for(int i=0; i<n; i++)
            dp[i]=false;
        for(int j=n-1; j>=0; j--)
        {
            for(int i=j; i<n; i++)
            {
                string t(s.begin()+j,s.begin()+i+1);
                if( dict.find(t)!=dict.end() )
                {
                    if( ( i+1<n && dp[i+1])|| i==n-1)
                        dp[j]=true;
                }
            }
        }
        dfs(0,n,s,string (""),dict,dp);
        return ss;
    }
    void dfs(int st,int n,string str,string cur,unordered_set<string> &dict,bool dp[])
    {
        if(st==n)
        {
            ss.push_back(cur);
            return ;
        }
        for(int i=st; i<n; i++)
        {
            string t(str.begin()+st,str.begin()+i+1);
            if((dp[i+1]||i+1==n) &&  (dict.find(t)!=dict.end() ))
            {
                int c=cur.size();
                cur+=t;
                if(i<n-1)
                    cur.push_back(' ');
                dfs(i+1,n,str,cur,dict,dp);
                cur.resize(c);
            }
        }
    }
};

int main()
{
    unordered_set<string> dict= {};
    Solution s;
    string ss = "a";
    vector<string> result=s.wordBreak(ss,dict);
    for(auto a:result)
        cout<<a<<endl;
}

参考：http://blog.csdn.net/cs_guoxiaozhu/article/details/14104789