970. Powerful Integers

#### QUESTION:

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

#### SOLUTION:

``````class Solution {
public List<Integer> powerfulIntegers(int x, int y, int bound) {
Set<Integer> seen = new HashSet<>();
for (int m = 1; m <= bound; m *= x) {
for (int n = 1; m + n <= bound; n *= y) {
if (y == 1) break;
}
if (x == 1) break;
}
return new ArrayList<>(seen);
}
}
//哈哈，我的方法比较丑陋，前面的结果只需要1ms，而我的需要4ms。
public static List<Integer> powerfulIntegers(int x, int y, int bound) {
double sum = 1;int tmp = 0;
List<Double> powX = new ArrayList<>();
if(x!=1){
while (sum<bound){
tmp++;
sum = Math.pow(x,tmp);
}
sum = 1;tmp = 0;
List<Double> powY = new ArrayList<>();
if(y!=1){
while (sum<bound){
tmp++;
sum = Math.pow(y,tmp);
}
HashSet<Double> set = new HashSet<>();
for(int i = 0;i<powX.size();i++){
in:for(int j = 0;j<powY.size();j++){
double res = powX.get(i)+powY.get(j);
if(res > bound) break in;