LeetCode 587 - Erect the Fence

http://bookshadow.com/weblog/2017/05/15/leetcode-erect-the-fence/

There are some trees, where each tree is represented by (x,y) coordinate in a two-dimensional garden. Your job is to fence the entire garden using the minimum length of rope as it is expensive. The garden is well fenced only if all the trees are enclosed. Your task is to help find the coordinates of trees which are exactly located on the fence perimeter.

Example 1:

Input: [[1,1],[2,2],[2,0],[2,4],[3,3],[4,2]]
Output: [[1,1],[2,0],[4,2],[3,3],[2,4]]
Explanation:

Example 2:

Input: [[1,2],[2,2],[4,2]]
Output: [[1,2],[2,2],[4,2]]
Explanation:

Even you only have trees in a line, you need to use rope to enclose them. 

Note:

1. All trees should be enclosed together. You cannot cut the rope to enclose trees that will separate them in more than one group.
2. All input integers will range from 0 to 100.
3. The garden has at least one tree.
4. All coordinates are distinct.
5. Input points have NO order. No order required for output.

http://blog.csdn.net/u014688145/article/details/71948763
public List<Point> outerTrees(Point[] points) { Point first = points[0]; int firstIndex = 0; // Find the leftmost point for (int i = 0; i < points.length; i++) { Point point = points[i]; if (point.x < first.x) { first = point; firstIndex = i; } } Set<Point> answer = new HashSet<>(); Point cur = first; int curIndex = firstIndex; answer.add(first); do { Point next = points[0]; int nextIndex = 0; for (int i = 1; i < points.length; i++) { if (i == curIndex) continue; Point p = points[i]; int cross = crossProductLength(p, cur, next); if (nextIndex == curIndex || cross > 0 || // Handle multi points in a line (cross == 0 && distance(p, cur) > distance(next, cur))) { next = p; nextIndex = i; } } // Handle multi points in a line for (int i = 0; i < points.length; i++) { Point p = points[i]; int cross = crossProductLength(p, cur, next); if (i != curIndex && cross == 0) { answer.add(p); } } cur = next; curIndex = nextIndex; } while (curIndex != firstIndex); return new ArrayList<>(answer); } private int crossProductLength(Point A, Point B, Point C) { // Get the vectors' coordinates. int BAx = A.x - B.x; int BAy = A.y - B.y; int BCx = C.x - B.x; int BCy = C.y - B.y; // Calculate the Z coordinate of the cross product. return (BAx * BCy - BAy * BCx); } private int distance(Point p1, Point p2) { return (p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y); }

凸包（Convex Hull）

Graham扫描法

1. 求点集points中的最左下点lb

2. 以lb为原点，对points按照极坐标排序

3. 维护栈stack，初始将points[0], points[1]压入栈

4. 从2到len(points) - 1遍历各点i

    重复弹出栈顶，直到points[-2], points[-1], points[i]非顺时针旋转未止

    将points[i]压入栈

需要注意的是，当最大极角对应的点超过一个时，按照到lb的距离逆序排列。