1276 Number of Burgers with No Waste of Ingredients 不浪费原料的汉堡制作方案

Description:

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:

Example 1:

Input: tomatoSlices = 16, cheeseSlices = 7
Output: [1,6]
Explantion: To make one jumbo burger and 6 small burgers we need 41 + 26 = 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.

Constraints:

0 <= tomatoSlices, cheeseSlices <= 10^7

题目描述：

圣诞活动预热开始啦，汉堡店推出了全新的汉堡套餐。为了避免浪费原料，请你帮他们制定合适的制作计划。

给你两个整数 tomatoSlices 和 cheeseSlices，分别表示番茄片和奶酪片的数目。不同汉堡的原料搭配如下：

巨无霸汉堡：4 片番茄和 1 片奶酪
小皇堡：2 片番茄和 1 片奶酪
请你以 [total_jumbo, total_small]（[巨无霸汉堡总数，小皇堡总数]）的格式返回恰当的制作方案，使得剩下的番茄片 tomatoSlices 和奶酪片 cheeseSlices 的数量都是 0。

如果无法使剩下的番茄片 tomatoSlices 和奶酪片 cheeseSlices 的数量为 0，就请返回 []。

示例：

示例 1：

输入：tomatoSlices = 16, cheeseSlices = 7
输出：[1,6]
解释：制作 1 个巨无霸汉堡和 6 个小皇堡需要 41 + 26 = 16 片番茄和 1 + 6 = 7 片奶酪。不会剩下原料。

示例 2：

输入：tomatoSlices = 17, cheeseSlices = 4
输出：[]
解释：只制作小皇堡和巨无霸汉堡无法用光全部原料。

示例 3：

输入：tomatoSlices = 4, cheeseSlices = 17
输出：[]
解释：制作 1 个巨无霸汉堡会剩下 16 片奶酪，制作 2 个小皇堡会剩下 15 片奶酪。

示例 4：

输入：tomatoSlices = 0, cheeseSlices = 0
输出：[0,0]

示例 5：

输入：tomatoSlices = 2, cheeseSlices = 1
输出：[0,1]

提示：

0 <= tomatoSlices <= 10^7
0 <= cheeseSlices <= 10^7

思路：

数学
可以列出二元一次方程,
设巨无霸汉堡个数为 x, 小皇堡个数为 y
4x + 2y = tomatoSlices
x + y = cheeseSlices
解方程可得
x = (tomatoSlices - cheeseSlices * 2) / 2
y = cheeseSlices - x
x, y 都必须为正整数
时间复杂度为 O(1), 空间复杂度为 O(1)

代码:

C++:

class Solution 
{
public:
    vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) 
    {
    return (tomatoSlices & 1 or tomatoSlices > (cheeseSlices << 2) or tomatoSlices < (cheeseSlices << 1)) ? vector<int>{} : vector<int>{(tomatoSlices - (cheeseSlices << 1) >> 1), cheeseSlices - (tomatoSlices - (cheeseSlices << 1) >> 1)};
    }
};

Java:

class Solution {
    public List<Integer> numOfBurgers(int tomatoSlices, int cheeseSlices) {
        return ((tomatoSlices & 1) != 0 || (tomatoSlices > (cheeseSlices << 2)) || tomatoSlices < (cheeseSlices << 1)) ? new ArrayList<>() : new ArrayList<>(){{ add((tomatoSlices - (cheeseSlices << 1) >> 1)); add(cheeseSlices - (tomatoSlices - (cheeseSlices << 1) >> 1)); }};
    }
}

Python:

class Solution:
    def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:
        return [] if tomatoSlices & 1 or tomatoSlices > (cheeseSlices << 2) or tomatoSlices < (cheeseSlices << 1) else [(a := tomatoSlices - (cheeseSlices << 1) >> 1), cheeseSlices - a]