leetcode1465. Maximum Area of a Piece of Cake After Horizontal and Vertical Cuts

Posted 2020-12-12 seyjs

题目如下：

Given a rectangular cake with height h and width w, and two arrays of integers horizontalCuts and verticalCuts where horizontalCuts[i] is the distance from the top of the rectangular cake to the ith horizontal cut and similarly, verticalCuts[j] is the distance from the left of the rectangular cake to the jth vertical cut.

Return the maximum area of a piece of cake after you cut at each horizontal and vertical position provided in the arrays horizontalCuts and verticalCuts. Since the answer can be a huge number, return this modulo 10^9 + 7.

Example 1:

Input: h = 5, w = 4, horizontalCuts = [1,2,4], verticalCuts = [1,3]
Output: 4 
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. 
After you cut the cake, the green piece of cake has the maximum area.

Example 2:

Input: h = 5, w = 4, horizontalCuts = [3,1], verticalCuts = [1]
Output: 6
Explanation: The figure above represents the given rectangular cake. Red lines are the horizontal and vertical cuts. After you cut the cake, the green and yellow pieces of cake have the maximum area.

Example 3:

Input: h = 5, w = 4, horizontalCuts = [3], verticalCuts = [3]
Output: 9

Constraints:

• 2 <= h, w <= 10^9
• 1 <= horizontalCuts.length < min(h, 10^5)
• 1 <= verticalCuts.length < min(w, 10^5)
• 1 <= horizontalCuts[i] < h
• 1 <= verticalCuts[i] < w
• It is guaranteed that all elements in horizontalCuts are distinct.
• It is guaranteed that all elements in verticalCuts are distinct.

解题思路：本题其实很简单，只要分别找出horizontalCuts和verticalCuts相邻的两个元素的最大差值，再计算这两个差值的成绩即可。

代码如下：

class Solution(object):
    def maxArea(self, h, w, horizontalCuts, verticalCuts):
        """
        :type h: int
        :type w: int
        :type horizontalCuts: List[int]
        :type verticalCuts: List[int]
        :rtype: int
        """
        horizontalCuts.sort()
        verticalCuts.sort()
        if horizontalCuts[0] != 0:horizontalCuts.insert(0,0)
        if horizontalCuts[-1] != h:horizontalCuts.append(h)

        if verticalCuts[0] != 0:verticalCuts.insert(0,0)
        if verticalCuts[-1] != w:verticalCuts.append(w)

        max_h,max_w = 0,0
        for i in range(1,len(horizontalCuts)):
            max_h = max(max_h,horizontalCuts[i] - horizontalCuts[i-1])

        for i in range(1,len(verticalCuts)):
            max_w = max(max_w,verticalCuts[i] - verticalCuts[i-1])

        return (max_h * max_w) % (10**9+7)