LeetCode 1274. 矩形内船只的数目（分治）

文章目录

- 1. 题目
- 2. 解题

1. 题目

(此题是交互式问题 )

在用笛卡尔坐标系表示的二维海平面上，有一些船。

每一艘船都在一个整数点上，且每一个整数点最多只有 1 艘船。

有一个函数 Sea.hasShips(topRight, bottomLeft) ，输入参数为右上角和左下角两个点的坐标，当且仅当这两个点所表示的矩形区域（包含边界）内至少有一艘船时，这个函数才返回 true ，否则返回 false 。

给你矩形的右上角 topRight 和左下角 bottomLeft 的坐标，请你返回此矩形内船只的数目。

题目保证矩形内至多只有 10 艘船。

调用函数 hasShips 超过400次 的提交将被判为 错误答案（Wrong Answer）。
同时，任何尝试绕过评测系统的行为都将被取消比赛资格。

示例：
在这里插入图片描述

输入：
ships = [[1,1],[2,2],[3,3],[5,5]], 
topRight = [4,4], bottomLeft = [0,0]
输出：3
解释：在 [0,0] 到 [4,4] 的范围内总共有 3 艘船。提示：
ships 数组只用于评测系统内部初始化。
你无法得知 ships 的信息，所以只能通过调用 hasShips 接口来求解。
0 <= bottomLeft[0] <= topRight[0] <= 1000
0 <= bottomLeft[1] <= topRight[1] <= 1000

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/number-of-ships-in-a-rectangle
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

计算横纵坐标的中点，将矩形分成4块。

/*** // This is Sea's API interface.* // You should not implement it, or speculate about its implementation* class Sea {*   public:*     bool hasShips(vector<int> topRight, vector<int> bottomLeft);* };*/class Solution { //C++int sum = 0;
public:int countShips(Sea sea, vector<int> topRight, vector<int> bottomLeft) {if(topRight[0] < bottomLeft[0] || topRight[1] < bottomLeft[1]|| !sea.hasShips(topRight, bottomLeft))return 0;if(topRight == bottomLeft)return ++sum;int xmid = (topRight[0] + bottomLeft[0])/2;int ymid = (topRight[1] + bottomLeft[1])/2;countShips(sea, {xmid, ymid}, bottomLeft);countShips(sea,  {topRight[0], ymid}, {xmid+1, bottomLeft[1]});countShips(sea, {xmid, topRight[1]}, {bottomLeft[0], ymid+1});countShips(sea, topRight, {xmid+1, ymid+1});return sum;}
};

# """
# This is Sea's API interface.
# You should not implement it, or speculate about its implementation
# """
#class Sea(object):
#    def hasShips(self, topRight: 'Point', bottomLeft: 'Point') -> bool:
#
#class Point(object):
#   def __init__(self, x: int, y: int):
#       self.x = x
#       self.y = yclass Solution(object): #py3def __init__(self):self.sum = 0def countShips(self, sea: 'Sea', topRight: 'Point', bottomLeft: 'Point') -> int:if topRight.x < bottomLeft.x or topRight.y < bottomLeft.y or not sea.hasShips(topRight, bottomLeft):return 0xmid = (topRight.x + bottomLeft.x)//2ymid = (topRight.y + bottomLeft.y)//2if topRight.x == bottomLeft.x and topRight.y == bottomLeft.y:self.sum += 1return self.sumself.countShips(sea, Point(xmid, ymid), bottomLeft)self.countShips(sea, Point(topRight.x, ymid), Point(xmid+1, bottomLeft.y))self.countShips(sea, Point(xmid, topRight.y), Point(bottomLeft.x, ymid+1))self.countShips(sea, topRight, Point(xmid+1, ymid+1))return self.sum