You are given an array of n integers, nums, where there are at most 50 unique values in the array. You are also given an array of m customer order quantities, quantity, where quantity[i] is the amount of integers the ith customer ordered. Determine if it is possible to distribute nums such that:

• The ith customer gets exactly quantity[i] integers,
• The integers the ith customer gets are all equal, and
• Every customer is satisfied.

Return true if it is possible to distribute nums according to the above conditions.

Example 1:

Input: nums = [1,2,3,4], quantity = [2] Output: false Explanation: The 0th customer cannot be given two different integers. 

Example 2:

Input: nums = [1,2,3,3], quantity = [2] Output: true Explanation: The 0th customer is given [3,3]. The integers [1,2] are not used. 

Example 3:

Input: nums = [1,1,2,2], quantity = [2,2] Output: true Explanation: The 0th customer is given [1,1], and the 1st customer is given [2,2]. 

Example 4:

Input: nums = [1,1,2,3], quantity = [2,2] Output: false Explanation: Although the 0th customer could be given [1,1], the 1st customer cannot be satisfied. 

Example 5:

Input: nums = [1,1,1,1,1], quantity = [2,3] Output: true Explanation: The 0th customer is given [1,1], and the 1st customer is given [1,1,1]. 

Constraints:

• n == nums.length
• 1 <= n <= 105
• 1 <= nums[i] <= 1000
• m == quantity.length
• 1 <= m <= 10
• 1 <= quantity[i] <= 105
• There are at most 50 unique values in nums.

给你一个长度为 n 的整数数组 nums ，这个数组中至多有 50 个不同的值。同时你有 m 个顾客的订单 quantity ，其中，整数 quantity[i] 是第 i 位顾客订单的数目。请你判断是否能将 nums 中的整数分配给这些顾客，且满足：

• 第 i 位顾客 恰好 有 quantity[i] 个整数。
• 第 i 位顾客拿到的整数都是 相同的 。
• 每位顾客都满足上述两个要求。

如果你可以分配 nums 中的整数满足上面的要求，那么请返回 true ，否则返回 false 。

• 给定一个数组 nums，订单数组 quantity，要求按照订单满足顾客的需求。如果能满足输出 true，不能满足输出 false。
• 用 DFS 记忆化暴力搜索。代码实现不难。（不知道此题为什么是 Hard）

package leetcode funccanDistribute(nums []int, quantity []int) bool { freq:=make(map[int]int) for_, n:=rangenums { freq[n]++ } returndfs(freq, quantity) } funcdfs(freqmap[int]int, quantity []int) bool { iflen(quantity) ==0 { returntrue } visited:=make(map[int]bool) fori:=rangefreq { ifvisited[freq[i]] { continue } visited[freq[i]] =trueiffreq[i] >=quantity[0] { freq[i] -=quantity[0] ifdfs(freq, quantity[1:]) { returntrue } freq[i] +=quantity[0] } } returnfalse }