LeetCode.970- powerful integer (Powerful Integers)

This is the first book of pleasure and delight 367 update, the first 395 Pian original

01 questions and look ready

Introduced today is the LeetCodearithmetic problem in Easythe first-level 229title (overall title number is 970). Given two positive integers xand y, if for some integer i >= 0and is j >= 0equal x^i + y^j, then the integer is powerful.

The return value is less than or equal to the boundlist of all the powerful integers.

You can return the answers in any order. In your answer, each value should appear at most once. E.g:

Input: x = 2, y = 3 , bound 10 =
Output: [2,3,4,5,7,9,10]
Description:
2 + 0 = 2 ^ 3 ^ 0
3 = 2 + 1 ^ 3 ^ 0
4 + 0 = 2 ^ 3 ^ 1
5 = 2 + 1 ^ 3 ^ 1
. 7 = 2 ^ 2 ^ 3 1 +
9 = 2 ^ 3 ^ 3 + 0
10 = 0 + 2 ^ 2 ^ 3

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
  
  

02 The first solution

Can be directly translated title, no special skills, but need to note that, because the judgment condition x or y is less than a few parties bound, if xor yas a time of 1, no power will be 1 1, would have been less than bound, It will result in an endless loop.

public List<Integer> powerfulIntegers(int x, int y, int bound) {
    Set<Integer> set = new HashSet<Integer>();
    for (int i=0; Math.pow(x, i) < bound; i++) {
        for (int j=0; Math.pow(y, j) < bound; j++) {
            int sum = (int)Math.pow(x, i)+(int)Math.pow(y, j);
            if (sum <= bound) {
                set.add((int)sum);
            }
            // y等于1时，容易造成死循环，要结束掉
            if (y == 1) {
                break;
            }
        }
        // x等于1时，容易造成死循环，要结束掉
        if (x == 1) {
            break;
        }
    }
    return new ArrayList<>(set);
}

03 The second solution

For the first solution above, we can not help Mathclass powmethod, multiplied by a total replacement, ideas are the same.

public List<Integer> powerfulIntegers2(int x, int y, int bound) {
    Set<Integer> set = new HashSet<Integer>();
    for (int i=1; i<bound; i *= x) {
        for (int j=1; j<bound; j *= y) {
            if (i+j <= bound) {
                set.add(i+j);    
            }
            if (y == 1) {
                break;
            }
        }
        if (x == 1) {
            break;
        }
    }
    return new ArrayList<>(set);
}

04 Summary

Thematic algorithm has been continuously day and more than seven months , the algorithm of feature articles 235 + articles, public Number dialog box reply [ data structures and algorithms ], [ algorithm ], [ data structures ] either a keyword to obtain a series of articles Collection .

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