[LeetCode] 970. Powerful Integers 强力数字
Given two positive integers x
and y
, an integer is powerful if it is equal to x^i + y^j
for some integers i >= 0
and j >= 0
.
Return a list of all powerful integers that have value less than or equal to bound
.
You may return the answer in any order. In your answer, each value should occur at most once.
Example 1:
Input: x = 2, y = 3, bound = 10
Output: [2,3,4,5,7,9,10]
Explanation:
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
Example 2:
Input: x = 3, y = 5, bound = 15
Output: [2,4,6,8,10,14]
Note:
1 <= x <= 100
1 <= y <= 100
0 <= bound <= 10^6
这道题定义了一种强力整数,说是给定的整数x和y分别的i次幂和j次幂之和,现在又给了一个整数 bound,让返回不超过这个范围的所有的强力整数。既然是一道 Easy 的题目,就不要考虑太多的技巧了,直接上无脑破解了吧。博主最开始的解法是在 bound 范围内先分别生成x和y的指数数组,即 x^0, x^1, x^2....
和 y^0, y^1, y^2....
,然后从两个数组中各自任意取出一个数字来相加,只要不超过 bound,就可以放入结果 res 中了,需要注意的是,若x和y等于1的话,那么会陷入死循环,因为乘以1永远等于其本身,所以要加另外的判断。博主的方法其实可以优化一下,没有必要用额外的数组去保存,而是可以直接在 for 循环中处理就可以了。还有,为了防止重复数字,先是把结果都存入一个 TreeSet 中,利用其可以去除重复项的特点,最后再转回数组就行了,参见代码如下:
class Solution {
public:
vector<int> powerfulIntegers(int x, int y, int bound) {
set<int> res;
for (int a = 1; a < bound; a *= x) {
for (int b = 1; a + b <= bound; b *= y) {
res.insert(a + b);
if (y == 1) break;
}
if (x == 1) break;
}
return vector<int>(res.begin(), res.end());
}
};
Github 同步地址:
https://github.com/grandyang/leetcode/issues/970
参考资料:
https://leetcode.com/problems/powerful-integers/
https://leetcode.com/problems/powerful-integers/discuss/214212/JavaC%2B%2BPython-Easy-Brute-Force