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:
- The given area won't exceed 10,000,000 and is a positive integer
- The web page's width and length you designed must be positive integers.
A:
---------------------瞧瞧自己有多傻------- 从sqrt开始,按照 L 和 W的距离越来越大 ,逐次寻找 . 就只第一次用了sqrt() 其实可以不用的 -------------------
class Solution { public: vector<int> constructRectangle(int area) { int W = sqrt(area); int dist = 0; vector<int> res; for(int dist = 0; dist < area; ++dist) { int t = W*(W+dist); if(t == area) { res = { W+dist, W}; break; }else if(t < area){ while(t<area){ ++W; t = W * (W+dist); } if(t == area){ res = {W+dist, W}; break; } }else{ // current is bigger while(t > area){ --W; t = W * (W+dist); } if(t == area){ res = {W+dist, W}; break; } } } return res; } };
---------------- 直接开始除------------------------TNND-------------
class Solution { public: vector<int> constructRectangle(int area) { for(int W = sqrt(area);W>=1;--W) { if(area%W ==0) return {area/W, W}; } return {-1,-1}; } };
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