【leetcode刷题】T55-强整数

输入：x = 2, y = 3, bound = 10
输出：[2,3,4,5,7,9,10]
解释： 
2 = 2^0 + 3^0
3 = 2^1 + 3^0
4 = 2^0 + 3^1
5 = 2^1 + 3^1
7 = 2^2 + 3^1
9 = 2^3 + 3^0
10 = 2^0 + 3^2

输入：x = 3, y = 5, bound = 15
输出：[2,4,6,8,10,14]

class Solution(object):
    def get_all_num(self, x, bound):
        ls = []
        if x == :
            return ls
        num = x
        while num < bound:
            ls.append(num)
            num *= x
        return ls
    
    def powerfulIntegers(self, x, y, bound):
        """
        :type x: int
        :type y: int
        :type bound: int
        :rtype: List[int]
        """
        ls1 = self.get_all_num(x, bound)
        ls2 = self.get_all_num(y, bound)
        res = []
        for ls1i in ls1:
            for ls2i in ls2:
                tmp = ls1i + ls2i
                if tmp > bound:
                    break
                res.append(tmp)
        return list(set(res))

class Solution {
public:
    vector<int> get_all_num(int x, int bound){
        vector<int> ls(,);
        if(x == )
            return ls;
        int num = x;
        while(num < bound){
            ls.push_back(num);
            num *= x;
        }
        return ls;
    }
    vector<int> powerfulIntegers(int x, int y, int bound) {
        vector<int> ls1 = get_all_num(x, bound);
        vector<int> ls2 = get_all_num(y, bound);
        map<int, int> d;
        int tmp;
        for(auto n1: ls1){
            for(auto n2:ls2){
               tmp = n1 + n2;
                if(tmp > bound)
                    break;
                d[tmp] = ;
            }
        }
        vector<int> res;
        for(map<int, int>::iterator it=d.begin(); it != d.end(); it++)
            res.push_back(it->first);
        return res;
    }
};