131. Palindrome Partitioning

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类似于subset

class Solution {
public:
    vector> partition(string s) {
        vector> result;
        vector list;
        if(s.empty()) return result;
        dfs(0, result, list, s);
        return result;
    }
private:
    void dfs(int pos, vector>& result, vector& list, string s){
        if(pos == s.size()){
            result.push_back(list);
        }
        for(int i = pos; i < s.size(); i++){//
            string substr = s.substr(pos, i + 1 - pos);
            if(!isPalindrome(substr)) continue;
            list.push_back(substr);
            dfs(i+1, result, list, s);//因为这种类似于拼接的题目，subset之类的，取过的不能取， 只能往后取了，所以要是i+1
            list.pop_back();
        }
    }
    bool isPalindrome(string str){
        if(str.empty()) return false;
        int l = 0, r = str.size() - 1;
        while(l <= r){
            if(str[l] != str[r]){
                return false;
            }
            l++;
            r--;
        }
        return true;
    }
};