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"].

代码如下：

class Solution {
    vector<string> midres;
    vector<string> res;
    vector<bool> *dp;
public:
    vector<string> wordBreak(string s, unordered_set<string> &dict) {
        int len = s.length();
        
        dp = new vector<bool>[len];
        for(int i=0; i<len; ++i){
            for(int j=i; j<len; ++j){
                if(dict.find(s.substr(i, j-i+1))!=dict.end()){
                    dp[i].push_back(true);      //第二维的下标实际是:单词长度-1
                }else{
                    dp[i].push_back(false);     //数组第二维用vector,size不一定是n,这样比n*n节省空间
                }
            }
        }
        func(s, len-1);
        return res;
    }
    
    void func(const string &s, int i){
        if(i>=0){
            for(int j=0; j<=i; ++j){
                
                if(dp[j][i-j]){ //注意此处的第二个下标是 i-j，不是i,因为数组的第二维长度是不固定的,第二维的下标实际是单词长度-1
                
                    midres.push_back(s.substr(j, i-j+1));
                    func(s, j-1);
                    midres.pop_back();  //继续考虑for循环的下一个分段处
                }
            }
            return;
        }
        else{
            string str;
            for(int k=midres.size()-1; k>=0; --k){  //注意遍历的顺序是倒序的
                str += midres[k];   //注意此处是k,不是i
                if(k>0)
                    str += " ";
            }
            res.push_back(str);
            return;
        }
    }
};