【leetcode】970. Powerful Integers

最新推荐文章于 2025-08-13 10:25:05 发布

weixin_34352005

最新推荐文章于 2025-08-13 10:25:05 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/seyjs/p/10233942.html

本文介绍了一种算法，用于找出所有小于等于特定界限的、由两个非负整数的幂次组合而成的强大整数。通过遍历可能的幂次组合并检查其是否满足条件，该算法有效地生成了所需整数列表。

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题目如下：

Given two non-negative 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

解题思路：注意，题目中的^不是异或而是幂。方法很简单，如果x/y等于1，那么幂值只会是1；如果x/y 大于1，由于 bound <= 10^6，幂的最大值是20(pow(2,20) > 10^6)。

代码如下：

class Solution(object):
    def powerfulIntegers(self, x, y, bound):
        """
        :type x: int
        :type y: int
        :type bound: int
        :rtype: List[int]
        """
        res = set()
        x_max,y_max = 20 if x > 1 else 1,20 if y > 1 else 1
        for i in range(x_max):
            for j in range(y_max):
                v = pow(x,i) + pow(y,j)
                if v <= bound:
                    res.add(v)
        return list(res)