[LeetCode] 1774. Closest Dessert Cost
作者:互联网
You would like to make dessert and are preparing to buy the ingredients. You have n
ice cream base flavors and m
types of toppings to choose from. You must follow these rules when making your dessert:
- There must be exactly one ice cream base.
- You can add one or more types of topping or have no toppings at all.
- There are at most two of each type of topping.
You are given three inputs:
baseCosts
, an integer array of lengthn
, where eachbaseCosts[i]
represents the price of theith
ice cream base flavor.toppingCosts
, an integer array of lengthm
, where eachtoppingCosts[i]
is the price of one of theith
topping.target
, an integer representing your target price for dessert.
You want to make a dessert with a total cost as close to target
as possible.
Return the closest possible cost of the dessert to target
. If there are multiple, return the lower one.
Example 1:
Input: baseCosts = [1,7], toppingCosts = [3,4], target = 10 Output: 10 Explanation: Consider the following combination (all 0-indexed): - Choose base 1: cost 7 - Take 1 of topping 0: cost 1 x 3 = 3 - Take 0 of topping 1: cost 0 x 4 = 0 Total: 7 + 3 + 0 = 10.
Example 2:
Input: baseCosts = [2,3], toppingCosts = [4,5,100], target = 18 Output: 17 Explanation: Consider the following combination (all 0-indexed): - Choose base 1: cost 3 - Take 1 of topping 0: cost 1 x 4 = 4 - Take 2 of topping 1: cost 2 x 5 = 10 - Take 0 of topping 2: cost 0 x 100 = 0 Total: 3 + 4 + 10 + 0 = 17. You cannot make a dessert with a total cost of 18.
Example 3:
Input: baseCosts = [3,10], toppingCosts = [2,5], target = 9 Output: 8 Explanation: It is possible to make desserts with cost 8 and 10. Return 8 as it is the lower cost.
Example 4:
Input: baseCosts = [10], toppingCosts = [1], target = 1 Output: 10 Explanation: Notice that you don't have to have any toppings, but you must have exactly one base.
Constraints:
n == baseCosts.length
m == toppingCosts.length
1 <= n, m <= 10
1 <= baseCosts[i], toppingCosts[i] <= 104
1 <= target <= 104
最接近目标价格的甜点成本。
你打算做甜点,现在需要购买配料。目前共有 n 种冰激凌基料和 m 种配料可供选购。而制作甜点需要遵循以下几条规则:
必须选择 一种 冰激凌基料。
可以添加 一种或多种 配料,也可以不添加任何配料。
每种类型的配料 最多两份 。
给你以下三个输入:baseCosts ,一个长度为 n 的整数数组,其中每个 baseCosts[i] 表示第 i 种冰激凌基料的价格。
toppingCosts,一个长度为 m 的整数数组,其中每个 toppingCosts[i] 表示 一份 第 i 种冰激凌配料的价格。
target ,一个整数,表示你制作甜点的目标价格。
你希望自己做的甜点总成本尽可能接近目标价格 target 。返回最接近 target 的甜点成本。如果有多种方案,返回 成本相对较低 的一种。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/closest-dessert-cost
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思路是DFS。这道题是在一定预算范围内找所有的可能性,所以有点类似于backtracking/permutation一类的题。我给出的解法也是类似。如果对permutation一类的题不熟悉可以先做题号比较小的题目。这道题跟一般求permutation类型的题目有点区别的地方就是对于toppings,可以不买,可以买一个,也可以买两个。
时间O(n)
空间O(n)
Java实现
1 class Solution { 2 int res = Integer.MAX_VALUE; 3 4 public int closestCost(int[] baseCosts, int[] toppingCosts, int target) { 5 for (int base : baseCosts) { 6 helper(base, toppingCosts, 0, target); 7 } 8 return res; 9 } 10 11 private void helper(int curCost, int[] toppingCosts, int index, int target) { 12 // 如果总价比res距离target更近 13 // 或者总价比target更小,则更新 14 if (Math.abs(curCost - target) < Math.abs(res - target) 15 || Math.abs(curCost - target) == Math.abs(res - target) && curCost < res) { 16 res = curCost; 17 } 18 // base case 19 if (index == toppingCosts.length) { 20 return; 21 } 22 helper(curCost, toppingCosts, index + 1, target); // 不买topping 23 helper(curCost + toppingCosts[index], toppingCosts, index + 1, target); // 买一个topping 24 helper(curCost + 2 * toppingCosts[index], toppingCosts, index + 1, target); // 买两个topping 25 } 26 }
标签:1774,topping,target,10,baseCosts,cost,toppingCosts,LeetCode,Cost 来源: https://www.cnblogs.com/cnoodle/p/14493375.html