Q:给定两条线段(表示为起点start = {X1, Y1}和终点end = {X2, Y2}),如果它们有交点,请计算其交点,没有交点则返回空值。
要求浮点型误差不超过10^-6。若有多个交点(线段重叠)则返回X值最小的点,X坐标相同则返回Y值最小的点。
示例 1:
输入:
line1 = {0, 0}, {1, 0}
line2 = {1, 1}, {0, -1}
输出: {0.5, 0}
示例 2:
输入:
line1 = {0, 0}, {3, 3}
line2 = {1, 1}, {2, 2}
输出: {1, 1}
示例 3:
输入:
line1 = {0, 0}, {1, 1}
line2 = {1, 0}, {2, 1}
输出: {},两条线段没有交点
提示:
坐标绝对值不会超过2^7
输入的坐标均是有效的二维坐标
A:
这个题……纯粹就是数学题。
参考:https://leetcode-cn.com/problems/intersection-lcci/solution/xiang-xi-fen-lei-tao-lun-he-bing-xiang-si-qing-kua/
public double[] intersection(int[] start1, int[] end1, int[] start2, int[] end2) {
double x = 0, y = 0;
double x1 = start1[0], y1 = start1[1],
x2 = end1[0], y2 = end1[1],
x3 = start2[0], y3 = start2[1],
x4 = end2[0], y4 = end2[1];
// 第一条直线平行于 X 轴
if (x1 - x2 == 0) {
// 第二条直线也平行于 X 轴
if (x3 - x4 == 0) {
// 两条线段有重叠:同一直线,且一个端点在另一条线段之内
if (x1 == x3 && Math.max(y1, y2) >= Math.min(y3, y4) && Math.min(y1, y2) <= Math.max(y3, y4))
return new double[] {x1, Math.max(Math.min(y1, y2), Math.min(y3, y4))};
// 另一条直线不平行于此直线
} else {
x = x1;
y = (x - x4) * (y3 - y4) / (x3 - x4) + y4;
}
// 第一条直线平行于 Y 轴
} else if (y1 - y2 == 0) {
// 四个端点的对称点
int[] newStart1 = new int[] {start1[1], start1[0]},
newEnd1 = new int[] {end1[1], end1[0]},
newStart2 = new int[] {start2[1], start2[0]},
newEnd2 = new int[] {end2[1], end2[0]};
double[] tmp = intersection(newStart1, newEnd1, newStart2, newEnd2);
return new double[] {tmp[1], tmp[0]};
// 第二条直线平行于 X 轴
} else if (x3 - x4 == 0)
return intersection(start2, end2, start1, end1);
// 第二条直线平行于 Y 轴
else if (y3 - y4 == 0)
return intersection(start2, end2, start1, end1);
// 两条直线平行,但不平行于 X 轴或 Y 轴
else if ((y1 - y2) / (x1 - x2) == (y3 - y4) / (x3 - x4) &&
x2 * (y2 - y1) / (x1 - x2) + y2 == x4 * (y4 - y3) / (x3 - x4) + y4) {
x = Math.max(Math.min(x1, x2), Math.min(x3, x4));
y = Math.max(Math.min(y1, y2), Math.min(y3, y4));
// 普通情况
} else {
x = ((y4 - y2) * (x1 - x2) * (x3 - x4) + x2 * (y1 - y2) * (x3 - x4) - x4 * (y3 - y4) * (x1 - x2))
/ ((y1 - y2) * (x3 - x4) - (y3 - y4) * (x1 - x2));
y = (x - x2) * (y1 - y2) / (x1 - x2) + y2;
}
// 交点可能存在的左、右、下、上边界
double left = Math.max(Math.min(x1, x2), Math.min(x3, x4)),
right = Math.min(Math.max(x1, x2), Math.max(x3, x4)),
bottom = Math.max(Math.min(y1, y2), Math.min(y3, y4)),
top = Math.min(Math.max(y1, y2), Math.max(y3, y4));
if (left <= x && x <= right && bottom <= y && y <= top)
return new double[] {x, y};
else return new double[] {};
}