【Lintcode】612 k closest points

在heap里就要排序，先按d排序，第二是x，第三是y？

1 维持一个大小为K的max-heap

2 然后后面每push一个进来，就把最长距离那个pop出去

3 所以push进heap的值是负的distance，因为到时候可以把最远距离的给pop出去

"""

Definition for a point.

class Point:

    def __init__(self, a=0, b=0):

        self.x = a

        self.y = b

"""

class Solution:

    """

    @param points: a list of points

    @param origin: a point

    @param k: An integer

    @return: the k closest points

    """

    def kClosest(self, points, origin, k):

        # write your code here

        import heapq

        heap = [(-self.distance(p,origin), p) for p in points[:k]]

        heapq.heapify(heap)

        ret = []

        for p in points[k:]:

            small_d, point = heapq.heappop(heap)

            heapq.heappush(heap,(small_d,point))

            d = self.distance(p, origin)

            if d>-small_d:

                continue

            elif d<-small_d:

                heapq.heappushpop(heap, (-d, p))

            else:

                if p.x

                    heapq.heapreplace(heap, (-d, p))

                elif p.y

                    heapq.heapreplace(heap, (-d, p))

        ret = [p for d, p in heap]

        ret.sort(key=lambda item: (self.distance(item, origin), item.x, item.y))

        return ret

    def distance(self, point, origin):

        return (point.x-origin.x)**2+(point.y-origin.y)**2