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  • 【leetcode】939. Minimum Area Rectangle

    题目如下:

    Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes.

    If there isn't any rectangle, return 0.

    Example 1:

    Input: [[1,1],[1,3],[3,1],[3,3],[2,2]]
    Output: 4
    

    Example 2:

    Input: [[1,1],[1,3],[3,1],[3,3],[4,1],[4,3]]
    Output: 2

    Note:

    1. 1 <= points.length <= 500
    2. 0 <= points[i][0] <= 40000
    3. 0 <= points[i][1] <= 40000
    4. All points are distinct.

    解题思路:题目中约定了矩形的各边要与X轴或者Y轴平行,所以不妨把所有的点按X轴进行分组。例如 [[1,1],[1,3],[3,1],[3,3],[2,2]],以X的坐标分组后得到 {1: [1, 3], 2: [2], 3: [1, 3]},接下来只要判断两个不同的X轴所对应的Y轴的list中是否存在相同的元素,如果相同则表示可以组成一个题目中要求的矩形。最后求出最小面积即可。

    代码如下:

    class Solution(object):
        def minAreaRect(self, points):
            """
            :type points: List[List[int]]
            :rtype: int
            """
            dic_x = {}
            res = float('inf')
            x_set = set()
            for (x,y) in points:
                dic_x[x] = dic_x.setdefault(x,[]) + [y]
                x_set.add(x)
            x_list = sorted(list(x_set))
    
            for kx1 in range(len(x_list)):
                for kx2 in range(kx1+1,len(x_list)):
                    intersection = sorted((list(set(dic_x[x_list[kx1]]) & set(dic_x[x_list[kx2]]))))
                    if len(intersection) < 2:
                        continue
                    minDis = intersection[-1] - intersection[0]
                    for i in range(len(intersection)-1):
                        minDis = min(minDis,intersection[i+1] - intersection[i])
                    #print kx1,kx2,intersection
                    res = min(res, minDis * abs(x_list[kx1]-x_list[kx2]))
            return res if res != float('inf') else 0
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10168101.html
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