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)
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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;
}
}
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Solution ( 5ms)
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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;
}
}
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Notes