Word Break II 字符串分割 动态规划+DFS

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"]

.

一开始直接用DFS来做，发现超时，思考了一下发现存在重复判断，比如第一次通过两个单词来到位置s[7]，DFS后面的字符串无解，第二次通过一个单词直接来到s[7]，此时又对后面的字符串进行了DFS，显然是多余的操作。

于是改用动态规划来记录是否有解，再用DFS得出结果。

/*2015.9.2cyq*/
#include <iostream>
#include <string>
#include <vector>
#include <unordered_set>
using namespace std;

//直接DFS，超时
class Solution1{
public:
vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
vector<string> result;
string path;
dfs(s,0,wordDict,path,result);
return result;
}
private:
void dfs(string &s,int start,unordered_set<string>& wordDict,string &path,vector<string> &result){
if(start==s.size()){
int m=path.size();
result.push_back(path);
result.back().resize(m-1);
return ;
}
for(int i=start;i<s.size();i++){
string tmp=s.substr(start,i-start+1);
if(wordDict.find(tmp)!=wordDict.end()){
path+=tmp+" ";
dfs(s,i+1,wordDict,path,result);
path.resize(path.size()-tmp.size()-1);
}
}
}
};

//动态规划DP
class Solution{
public:
vector<string> wordBreak(string s, unordered_set<string>& wordDict) {
int n=s.size();
vector<bool> canSolve(n+1,false);//canSolve[i]表示s[i,n-1]可分割成若干个单词
canSolve
=true;
//f[i][j]表示s[i,j]是一个字典中的单词且canSolve[j+1]=true;
vector<vector<bool>> f(n,vector<bool>(n,false));
for(int i=n-1;i>=0;i--){
for(int j=i;j<n;j++){
if(wordDict.find(s.substr(i,j-i+1))!=wordDict.end()&&canSolve[j+1]){
f[i][j]=true;
canSolve[i]=true;
}
}
}
vector<string> result;
string path;
dfs(s,0,f,canSolve,path,result);
return result;
}
private:
void dfs(string &s,int start,vector<vector<bool>> &f,vector<bool> &canSolve,string &path,vector<string> &result){
if(start==s.size()){
int m=path.size();
result.push_back(path);
result.back().resize(m-1);
return ;
}
if(!canSolve[start])
return ;//剪枝
for(int i=start;i<s.size();i++){
if(f[start][i]){
string tmp=s.substr(start,i-start+1);
path+=tmp+" ";
dfs(s,i+1,f,canSolve,path,result);
path.resize(path.size()-tmp.size()-1);
}
}
}
};

int main(){
string a[]={"cat","cats","and","sand","dog"};
vector<string> dict(a,a+sizeof(a)/sizeof(string));
unordered_set<string> wordDict;
for(int i=0;i<dict.size();i++)
wordDict.insert(dict[i]);
Solution solu;
string s="catsanddog";
vector<string> ret=solu.wordBreak(s,wordDict);
for(auto it=ret.begin();it!=ret.end();++it){
cout<<*it<<endl;
}
return 0;
}