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An integer N is given, representing the area of some rectangle.
The area of a rectangle whose sides are of length A and B is A * B, and theperimeter is 2 * (A + B).
The goal is to find the minimal perimeter of any rectangle whose area equals N. The sides of this rectangle should be only integers.
For example, given integer N = 30, rectangles of area 30 are:
- (1, 30), with a perimeter of 62,
- (2, 15), with a perimeter of 34,
- (3, 10), with a perimeter of 26,
- (5, 6), with a perimeter of 22.
Write a function:
class Solution { public int solution(int N); }
that, given an integer N, returns the minimal perimeter of any rectangle whose area is exactly equal to N.
For example, given an integer N = 30, the function should return 22, as explained above.
Assume that:
- N is an integer within the range [1..1,000,000,000].
Complexity:
- expected worst-case time complexity is O(sqrt(N));
- expected worst-case space complexity is O(1).
// you can also use imports, for example:
// import java.util.*;
// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(int N) {
// write your code in Java SE 8
int max_perimeter = 2*(1+N);
for(int i=2; i < Math.sqrt(N)+1; i++) {
if(N%i == 0) {
max_perimeter = Math.min(max_perimeter, 2*(i+N/i));
}
}
return max_perimeter;
}
}
https://codility.com/demo/results/training8VZXU6-PUT/Codlility---MinPerimeterRectangle
标签：title href ams white for bsp tps rip any
原文地址：http://www.cnblogs.com/samo/p/6991222.html
