[LeetCode] 492. Construct the Rectangle

轉自LeetCode

For a web developer, it is very important to know how to design a web page's size. So, given a specific rectangular web page’s area, your job by now is to design a rectangular web page, whose length L and width W satisfy the following requirements:

1. The area of the rectangular web page you designed must equal to the given target area.

2. The width W should not be larger than the length L, which means L >= W.

3. The difference between length L and width W should be as small as possible.

You need to output the length L and the width W of the web page you designed in sequence.

Example:

Input: 4
Output: [2, 2]
Explanation: The target area is 4, and all the possible ways to construct it are [1,4], [2,2], [4,1]. 
But according to requirement 2, [1,4] is illegal; according to requirement 3,  [4,1] is not optimal compared to [2,2]. So the length L is 2, and the width W is 2.

Note:

1. The given area won't exceed 10,000,000 and is a positive integer
2. The web page's width and length you designed must be positive integers.

<Solution>

想法如下

• 因為第二個限制條件，可以知道 W <= sqrt(area)，不然的話，L就會小於W

• 從 W = sqrt(area) 往下找，找到可以將 area 整除的 W 就是答案了

code 如下(參考解法)

C++

	class Solution {
	public:
	vector<int> constructRectangle(int area) {
	vector<int> ans(2);
	//>> width's upperbound in sqrt(area)
	int width = static_cast<int>(sqrt((float)area));
	//>> find length
	while(area % width != 0) {
	--width;
	}
	ans[0] = area/width;
	ans[1] = width;
	return ans;
	}
	};

view raw constructRectangle.cpp hosted with ❤ by GitHub

Java

	class Solution {
	public int[] constructRectangle(int area) {
	int width = (int) Math.sqrt(area);
	while(area % width != 0) {
	--width;
	}
	return new int[] {area/width, width};
	}
	}

view raw constructRectangle.java hosted with ❤ by GitHub