05 August 2008
给定 X-Y 平面中的点数组，其中 points[i] = [xi, yi]。
返回由这些点形成的任何矩形的最小面积，其边不一定平行于 X 和 Y 轴。如果不存在这样的矩形，则返回 0。
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class Solution {
public double minAreaFreeRect(int[][] points) {
if (points.length < 4) {
return 0;
}
Set<String> set = new HashSet<>();
for (int i = 0; i < points.length; i++) {
set.add(format().apply(points[i]));
}
double area = 0;
for (int i = 0; i < points.length; i++) {
for (int j = i + 1; j < points.length; j++) {
int d2ij = distance2(points[i], points[j]);
for (int k = j + 1; k < points.length; k++) {
int d2ik = distance2(points[i], points[k]);
int d2jk = distance2(points[j], points[k]);
if (d2ij + d2ik != d2jk) {
continue;
}
int x = points[j][0] + points[k][0] - points[i][0];
int y = points[j][1] + points[k][1] - points[i][1];
int[] pointXY = new int[]{x, y};
if (!set.contains(format().apply(pointXY))) {
continue;
}
double tmpArea = Math.sqrt(d2ij) * Math.sqrt(d2ik);
if (area == 0 || area > tmpArea) {
area = tmpArea;
}
}
}
}
return area;
}
private Function<int[], String> format() {
return point -> String.format("%d_%d", point[0], point[1]);
}
private int distance2(int[] a, int[] b) {
return (b[0] - a[0]) * (b[0] - a[0]) + (b[1] - a[1]) * (b[1] - a[1]);
}
}