【leetcode】1276. Number of Burgers with No Waste of Ingredients

题目如下：

Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:

Jumbo Burger: 4 tomato slices and 1 cheese slice.

Small Burger: 2 Tomato slices and 1 cheese slice.

Return [total_jumbo, total_small] so that the number of remaining tomatoSlices equal to 0 and the number of remaining cheeseSlices equal to 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return [].

Example 1:
Input: tomatoSlices = 16, cheeseSlices = 7

Output: [1,6]

Explantion: To make one jumbo burger and 6 small burgers we need 4*1 + 2*6 = 16 tomato and 1 + 6 = 7 cheese. There will be no remaining ingredients.
Example 2:
Input: tomatoSlices = 17, cheeseSlices = 4

Output: []

Explantion: There will be no way to use all ingredients to make small and jumbo burgers.
Example 3:
Input: tomatoSlices = 4, cheeseSlices = 17

Output: []

Explantion: Making 1 jumbo burger there will be 16 cheese remaining and making 2 small burgers there will be 15 cheese remaining.
Example 4:
Input: tomatoSlices = 0, cheeseSlices = 0

Output: [0,0]
Example 5:
Input: tomatoSlices = 2, cheeseSlices = 1

Output: [0,1]
Constraints:

0 <= tomatoSlices <= 10^7

0 <= cheeseSlices <= 10^7

解题思路：解二元一次方程。

代码如下：

class Solution(object):

    def numOfBurgers(self, tomatoSlices, cheeseSlices):

        """

        :type tomatoSlices: int

        :type cheeseSlices: int

        :rtype: List[int]

        """

        if (tomatoSlices - 2*cheeseSlices)%2 != 0 or (4*cheeseSlices - tomatoSlices) % 2 != 0:

            return []

        elif (tomatoSlices - 2*cheeseSlices)/2 < 0 or (4*cheeseSlices - tomatoSlices) / 2 < 0:

            return []

        return [(tomatoSlices - 2*cheeseSlices)/2, (4*cheeseSlices - tomatoSlices) / 2]

巴特西

【leetcode】1276. Number of Burgers with No Waste of Ingredients

最新文章

热门文章