1. 强整数 - Powerful Integers

Problem Link

给定两个非负整数 x 和 y,如果某一整数等于 x^i + y^j,其中整数 i >= 0 且 j >= 0,那么我们认为该整数是一个强整数。

返回值小于或等于 bound 的所有强整数组成的列表。

你可以按任何顺序返回答案。在你的回答中,每个值最多出现一次。

Example:

示例 1:

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

示例 2:

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

Solutions

Solution 【BF】 ( 22ms)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution {
	public List<Integer> powerfulIntegers(int x, int y, int bound) {
        boolean[] nums = new boolean[bound+1];
        List<Integer> result = new LinkedList<>();
        for (int i = 0; i <=  bound; i++) {
        	if(Math.pow(x, i)>bound)
        		break;
        	for (int j = 0; j < bound; j++) {
        		if(Math.pow(x, i) + Math.pow(y, j)>bound)
            		break;
        		nums[(int)(Math.pow(x, i) + Math.pow(y, j))] = true;
			}
		}
        for (int i = 1; i < nums.length; i++) {
			if(nums[i])
				result.add(i);
		}
        return result;
    }
}

Solution ( 5ms)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
class Solution {
    public List<Integer> powerfulIntegers(int x, int y, int bound) {
        List<Integer> list  = new ArrayList<Integer>();
        if (x==1&&y==1)
        {
            if(2<=bound)
                list.add(2);
            return list;
        }
        if(x==1||y==1){
            int t =0;
            while(Math.pow(x,t) + Math.pow(y,t)<=bound){
                list.add((int)(Math.pow(x,t) + Math.pow(y,t)));
                t++;
            }
            return list;
        }
        for(int i =0 ; Math.pow(x,i)<bound;i++){
            for(int j= 0;Math.pow(y,j)<bound;j++){
                int temp = (int)(Math.pow(x,i) + Math.pow(y,j));
                if(temp<=bound && !list.contains(temp))
                    list.add(temp);
            }
        }
        return list;
    }
}

Notes

updatedupdated2021-03-052021-03-05