【问题描述】面试题 16.03. 交点
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。【解答思路】
https://leetcode-cn.com/problems/intersection-lcci/solution/jiao-dian-by-leetcode-solution/
需要认真考虑所有情况！
联立方程各种优化！
1. 时间复杂度：O(1) 空间复杂度：O(1)
class Solution {public double[] intersection(int[] start1, int[] end1, int[] start2, int[] end2) {int x1 = start1[0], y1 = start1[1];int x2 = end1[0], y2 = end1[1];int x3 = start2[0], y3 = start2[1];int x4 = end2[0], y4 = end2[1];double[] ans = new double[2];Arrays.fill(ans, Double.MAX_VALUE);// 判断两直线是否平行if ((y4-y3)*(x2-x1) == (y2-y1)*(x4-x3)) {// 判断两直线是否重叠if ((y2-y1)*(x3-x1) == (y3-y1)*(x2-x1)) {// 判断 (x3, y3) 是否在「线段」(x1, y1)~(x2, y2) 上if (isInside(x1, y1, x2, y2, x3, y3)) {updateRes(ans, x3, y3);}// 判断 (x4, y4) 是否在「线段」(x1, y1)~(x2, y2) 上if (isInside(x1, y1, x2, y2, x4, y4)) {updateRes(ans, (double)x4, (double)y4);}// 判断 (x1, y1) 是否在「线段」(x3, y3)~(x4, y4) 上if (isInside(x3, y3, x4, y4, x1, y1)) {updateRes(ans, (double)x1, (double)y1);}// 判断 (x2, y2) 是否在「线段」(x3, y3)~(x4, y4) 上if (isInside(x3, y3, x4, y4, x2, y2)) {updateRes(ans, (double)x2, (double)y2);}}} else {// 联立方程得到 t1 和 t2 的值double t1 = (double)(x3 * (y4 - y3) + y1 * (x4 - x3) - y3 * (x4 - x3) - x1 * (y4 - y3)) / ((x2 - x1) * (y4 - y3) - (x4 - x3) * (y2 - y1));double t2 = (double)(x1 * (y2 - y1) + y3 * (x2 - x1) - y1 * (x2 - x1) - x3 * (y2 - y1)) / ((x4 - x3) * (y2 - y1) - (x2 - x1) * (y4 - y3));// 判断 t1 和 t2 是否均在 [0, 1] 之间if (t1 >= 0.0 && t1 <= 1.0 && t2 >= 0.0 && t2 <= 1.0) {ans[0] = x1 + t1 * (x2 - x1);ans[1] = y1 + t1 * (y2 - y1);}}if (ans[0] == Double.MAX_VALUE) {return new double[0];}return ans;}// 判断 (x, y) 是否在「线段」(x1, y1)~(x2, y2) 上// 这里的前提是 (x, y) 一定在「直线」(x1, y1)~(x2, y2) 上private boolean isInside(int x1, int y1, int x2, int y2, int x, int y) {// 若与 x 轴平行，只需要判断 x 的部分// 若与 y 轴平行，只需要判断 y 的部分// 若为普通线段，则都要判断return (x1 == x2 || (Math.min(x1, x2) <= x && x <= Math.max(x1, x2)))&& (y1 == y2 || (Math.min(y1, y2) <= y && y <= Math.max(y1, y2)));}private void updateRes(double[] ans, double x, double y) {if (x < ans[0] || (x == ans[0] && y < ans[1])) {ans[0] = x;ans[1] = y;}}
}
2. 时间复杂度：O(N) 空间复杂度：O(1)
