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package class50;
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import java.util.Arrays;
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public class Problem_0587_ErectTheFence {
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public static int[][] outerTrees(int[][] points) {
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int n = points.length;
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int s = 0;
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int[][] stack = new int[n << 1][];
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// x小的排前面，x一样的，y小的排前面
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Arrays.sort(points, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]);
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for (int i = 0; i < n; i++) {
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while (s > 1 && cross(stack[s - 2], stack[s - 1], points[i]) > 0) {
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s--;
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}
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stack[s++] = points[i];
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}
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for (int i = n - 2; i >= 0; i--) {
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while (s > 1 && cross(stack[s - 2], stack[s - 1], points[i]) > 0) {
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s--;
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}
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stack[s++] = points[i];
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}
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// 去重返回
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Arrays.sort(stack, 0, s, (a, b) -> b[0] == a[0] ? b[1] - a[1] : b[0] - a[0]);
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n = 1;
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for (int i = 1; i < s; i++) {
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// 如果i点，x和y，与i-1点，x和y都一样
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// i点与i-1点，在同一个位置，此时，i点不保留
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if (stack[i][0] != stack[i - 1][0] || stack[i][1] != stack[i - 1][1]) {
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stack[n++] = stack[i];
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}
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}
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return Arrays.copyOf(stack, n);
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}
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// 叉乘的实现
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// 假设有a、b、c三个点，并且给出每个点的(x,y)位置
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// 从a到c的向量，在从a到b的向量的哪一侧？
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// 如果a到c的向量，在从a到b的向量右侧，返回正数
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// 如果a到c的向量，在从a到b的向量左侧，返回负数
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// 如果a到c的向量，和从a到b的向量重合，返回0
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public static int cross(int[] a, int[] b, int[] c) {
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return (b[1] - a[1]) * (c[0] - b[0]) - (b[0] - a[0]) * (c[1] - b[1]);
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}
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public static void main(String[] args) {
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int[] a = { 4, 4 };
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int[] b = { 1, 1 };
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int[] c = { 1, 5 };
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System.out.println(cross(a, b, c));
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}
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}
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