leetcode 850. Rectangle Area II

给定一些矩形2 求覆盖面积 矩形不超过200个

1#

算法1 朴素思想 虽然朴素但是代码却有意思

利用容斥原理

复杂度高达 N*2^N

class Solution:
    def intersect(rec1,rec2):
        return [max(rec1[1],rec2[1]),
                    max(rec1[2],rec2[2]),
                    min(rec1[3],rec2[3]),
                    min(rec1[4],rec2[4]
                    ]
#这里是两个矩形的两点式求 矩形的相交子矩形 *非常值得思考

    def area(rec):
        dx=max(0,rec[2]-rec[0])
        dy=max(0,rec[3]-rec[1])
        return dx*dy

    ans=0

    for size in range(1,len(rectangles)+1):

        for group in itertools.combinations(rectangles,size):
            ans = ans +(-1)** (size+1) * area(reduce(intersect,group))
    return ans%mod

2#

点位压缩，压缩后进行暴力循环

同时压缩x和y

最后返回

class Solution(object):
    def rectangleArea(self, rectangles):
        N = len(rectangles)
        Xvals, Yvals = set(), set()
        for x1, y1, x2, y2 in rectangles:
            Xvals.add(x1); Xvals.add(x2)
            Yvals.add(y1); Yvals.add(y2)

        imapx = sorted(Xvals)
        imapy = sorted(Yvals)
        mapx = {x: i for i, x in enumerate(imapx)}
        mapy = {y: i for i, y in enumerate(imapy)}

        grid = [[0] * len(imapy) for _ in imapx]
        for x1, y1, x2, y2 in rectangles:
            for x in xrange(mapx[x1], mapx[x2]):
                for y in xrange(mapy[y1], mapy[y2]):
                    grid[x][y] = 1

        ans = 0
        for x, row in enumerate(grid):
            for y, val in enumerate(row):
                if val:
                    ans += (imapx[x+1] - imapx[x]) * (imapy[y+1] - imapy[y])
        return ans % (10**9 + 7)

N^3

3#

算法3 扫描线算法

将每一个矩形看作一个 "事件" 这样的事件

class Solution(object):
    def rectangleArea(self, rectangles):
        # Populate events
        OPEN, CLOSE = 0, 1
        events = []
        for x1, y1, x2, y2 in rectangles:
            events.append((y1, OPEN, x1, x2))
            events.append((y2, CLOSE, x1, x2))
        events.sort()

        def query():
            ans = 0
            cur = -1
            for x1, x2 in active:
                cur = max(cur, x1)
                ans += max(0, x2 - cur)
                cur = max(cur, x2)
            return ans

        active = []
        cur_y = events[0][0]
        ans = 0
        for y, typ, x1, x2 in events:
            # For all vertical ground covered, update answer
            ans += query() * (y - cur_y)

            # Update active intervals
            if typ is OPEN:
                active.append((x1, x2))
                active.sort()
            else:
                active.remove((x1, x2))

            cur_y = y

        return ans % (10**9 + 7)

4#

注意到刚才的3算法中使用了 区间维护的算法 这里使用线段树维护这个区间

使得达到 NlogN

下面是py 实现线段树

class Node:
    def __init__(self,start,end):
        self.start=start
        self.end=end
        self.mid=(start+end)//2

        self.active_count=0
        self.totle =0

        self._left=None
        self._right=None
    @property
    def right(self):
        self._right= self._right or Node(self.mid,self.end)
        return self._right

    @property
    def left(self):
        self._left=self._left or Node(self.start,self.mid)
        return self._left

    #更新 i j 合适的区域 + val
    #同时返回 i j 之间的x大小
    def update(self,i,j,val):
        print(str(i)+" "+str(j))
        if(i>=j):
            return 0

        if(i==self.start and j==self.end):
            self.active_count = self.active_count + val
        else:
            self.left .update( i,  min( self.mid ,j )  , val )
            self.right.update( max(self.mid,i)       ,j, val ) 

        #当前区域有 至少一个覆盖
        if(self.active_count>0):

            self.totle= X[self.end]-X[self.start]
        else:

            self.totle= self.left.totle + self.right.totle 

        return self.totle 

class Solution:
    def rectangleArea(self, rectangles):
        """
        :type rectangles: List[List[int]]
        :rtype: int
        """
        ACTIVE = 1
        DEACTIVE = -1
        global X
        X=set()
        events=[]
        for rect in rectangles:
            X.add(rect[0])
            X.add(rect[2])

            events.append([rect[1],rect[0],rect[2],ACTIVE])
            events.append([rect[3],rect[0],rect[2],DEACTIVE])

        X=sorted(X)
        events=sorted(events)
        pos2idx={ x:i for i,x in enumerate(X) }        

        sum_area=0
        y_cur=0
        y_cur_next=0
        x_cur=0

        SegNode = Node(0,len(pos2idx))
        print(pos2idx)
        for event in events:
            y_cur_next=event[0]
            sum_area=sum_area+(y_cur_next-y_cur)*x_cur

            x_cur=SegNode.update(pos2idx[event[1]],pos2idx[event[2]],event[3])
            print(event)
            print(x_cur)

            y_cur=y_cur_next

        return sum_area%(1000000000 + 7)

注意这里的 下标实际意义不是 容器 而是 标志

所以 会有

start mid

mid end

这样的划分方法 应该注意

另外利用python 的 property 很方便的写出了懒申请策略

(python 做点集压缩真的方便

付：

leetcode 56. Merge Intervals On 求overlap