Minimum Area Rectangle

update Dec 30 18:03, 2018

LeetCode

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.

Basic Idea:

因为只考虑平行于坐标轴的矩形，这种情况下一条对角线就可以确定一个矩形。我们只需要枚举所有对角线，检查另外两点是否存在，然后就可以计算矩形面积。时间复杂度 O(n^2)，因为共有 n(n-1)/2 个连线。

• Java Code:

    class Solution {
        public int minAreaRect(int[][] points) {
            Set<Integer> set = new HashSet<>();
            for (int[] point : points) {
                set.add(point[0] * 40001 + point[1]);
            }
            int minArea = Integer.MAX_VALUE;
            // 枚举所有连线，因为只考虑平行于坐标轴的矩形，一条对角线就可以确定一个矩形
            for (int i = 0; i < points.length; ++i) {
                for (int j = i + 1; j < points.length; ++j) {
                    int[] p1 = points[i], p2 = points[j];
                    if (p1[0] == p2[0] || p1[1] == p2[1]) continue;
                    int[] p3 = new int[]{p1[0], p2[1]};
                    int[] p4 = new int[]{p2[0], p1[1]};
                    if (set.contains(p3[0] * 40001 + p3[1]) && set.contains(p4[0] * 40001 + p4[1])) {
                        // 说明找到一个矩形，计算面积
                        minArea = Math.min(minArea, Math.abs(p1[0] - p2[0]) * Math.abs(p1[1] - p2[1]));
                    }
                }
            }
            return minArea < Integer.MAX_VALUE ? minArea : 0;
        }
    }