LintCode 612 - K Closest Points

https://xuezhashuati.blogspot.com/2017/03/lintcode-612-k-closest-points.html
Given some points and a point origin in two dimensional space, find k points out of the some points which are nearest to origin.
Return these points sorted by distance, if they are same with distance, sorted by x-axis, otherwise sorted by y-axis.


Example
Given points = [[4,6],[4,7],[4,4],[2,5],[1,1]], origin = [0, 0], k = 3
return [[1,1],[2,5],[4,4]]


思路
根据题意，我们维护一个大小为K的max-heap。一个一个把point放进去，如果容量超了，就把最大的踢掉。这样heap里永远是最小的K个。（注意不是min-heap，自己举个例子就明白了。如果heap是[3, 4, 5]满了又来了2怎么办？当然是把5踢了，所以是max-heap。）
Comparator写的时候根据题意，如果距离相等，就比x轴，如果还相等，就比y轴。
最后把max-heap里面这K个points倒出来就是最近的K个。
另外算距离的时候不用开根，因为我们只比大小，所以勾方股方相加的数就够了可以比了。

    public Point globalOrigin = null;
    public Point[] kClosest(Point[] points, Point origin, int k) {        
        globalOrigin = origin;
        PriorityQueue<Point> pq = new PriorityQueue<Point>(k, new Comparator<Point>() {
            public int compare(Point a, Point b) {
                int diff = getDistance(b, globalOrigin) - getDistance(a, globalOrigin);
                if(diff == 0) {
                    diff = b.x - a.x;
                }
                if (diff == 0) {
                    diff = b.y - a.y;
                }
                return diff;
            }
        });
        
        for (Point pt : points) {
            pq.add(pt);
            if (pq.size() > k)
                pq.poll();
        }
        
        
        Point[] result = new Point[k];
        while (k - 1 >= 0) {
            result[--k] = pq.poll();
        }
        
        return result;
    }
    
    public int getDistance(Point a, Point b) {
        return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y); 
    }

https://liuyang89116.gitbooks.io/my-leetcode-book/content/k_closest_points.html

    public List<Point> findKClosestPoint(Point[] p, int k) {
        PriorityQueue<Point> pq = new PriorityQueue<Point>(k,
                new Comparator<Point>() {
                    public int compare(Point p1, Point p2) {
                        return (p2.x * p2.x + p2.y * p2.y)
                                - (p1.x * p1.x + p1.y * p1.y);
                    }
                });

        for (int i = 0; i < p.length; i++) {
            if (i < k) {
                pq.offer(p[i]);
            } else {
                Point curr = pq.peek();
                int d = p[i].x * p[i].x + p[i].y * p[i].y;
                int dCurr = curr.x * curr.x + curr.y * curr.y;
                if (d < dCurr) {
                    pq.poll();
                    pq.offer(p[i]);
                }
            }
        }

        List<Point> rst = new ArrayList<Point>();
        while (!pq.isEmpty()) {
            rst.add(pq.poll());
        }

        return rst;
    }

http://www.cnblogs.com/aprilyang/p/6702162.html

精度真是要注意再注意啊。。。

另外其实不用开根号的，这样就不会有损失精度的问题了。。。因为开根号可能不一样的减成0，然后又乱排了。