lc3218. 切蛋糕的最小总开销 i

题目链接：3218. 切蛋糕的最小总开销 I

题解

LC3219. 切蛋糕的最小总开销 II 的简化版本

参考代码

class Solution:
    def minimumCost(self, m: int, n: int, horizontalCut: List[int], verticalCut: List[int]) -> int:
    
        horizontalCut.sort()
        verticalCut.sort() 
        cost = 0
        i = 0
        j = 0
        while True:
            if m == 1 and n == 1:
                break
            # print(i, j, m, n)
            if n == 1:
                cost += horizontalCut[i]
                i += 1
                m -= 1
            elif m == 1:
                cost += verticalCut[j]
                j += 1
                n -= 1
            elif horizontalCut[i] < verticalCut[j]:
                cost += horizontalCut[i] * n
                i += 1
                m -= 1
            else:
                cost += verticalCut[j] * m
                j += 1
                n -= 1
            
        return cost