[Swift]LeetCode805. 数组的均值分割 | Split Array With Same Average

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In a given integer array A, we must move every element of A to either list B or list C. (B and C initially start empty.)

Return true if and only if after such a move, it is possible that the average value of B is equal to the average value of C, and B and C are both non-empty.

Example :
Input: 
[1,2,3,4,5,6,7,8]
Output: true
Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have the average of 4.5.

Note:

• The length of A will be in the range [1, 30].
• A[i] will be in the range of [0, 10000].

---

给定的整数数组 A ，我们要将 A数组 中的每个元素移动到 B数组 或者 C数组中。（B数组和C数组在开始的时候都为空）

返回true ，当且仅当在我们的完成这样的移动后，可使得B数组的平均值和C数组的平均值相等，并且B数组和C数组都不为空。

示例:
输入: 
[1,2,3,4,5,6,7,8]
输出: true
解释: 我们可以将数组分割为 [1,4,5,8] 和 [2,3,6,7], 他们的平均值都是4.5。

注意:

• A 数组的长度范围为 [1, 30].
• A[i] 的数据范围为 [0, 10000].

---

20ms

class Solution {
    func splitArraySameAverage(_ nums: [Int]) -> Bool {
        let sum = nums.reduce(0, +)
        let nums = nums.sorted()
        let n = nums.count
        if n < 2 { return false }
        for i in 1...n / 2 where (sum * i) % n == 0 && dfs(nums, i, 0, sum * i / n) { return true }
        return false
    }
    
    func dfs(_ nums: [Int], _ len: Int, _ idx: Int, _ sum: Int) -> Bool {
        if sum == 0, len == 0 { return true }
        if idx >= nums.count || len < 0 || sum < 0 { return false }
        for i in idx..<nums.count {
            if i > idx && nums[i] == nums[i - 1] { continue }
            if dfs(nums, len - 1, i + 1, sum - nums[i]) { return true }
        }
        return false
    }
}

---

Runtime: 604 ms

Memory Usage: 24.9 MB

class Solution {
    func splitArraySameAverage(_ A: [Int]) -> Bool {
        if A.count < 2 {return false}
        var sum:Int = A.reduce(0,+)
        var n:Int = A.count
        var m:Int = n / 2
        var possible:Bool = false
        var i:Int = 0
        while(i <= m && !possible)
        {
            if sum * i % n == 0 {possible = true}
            i += 1
        }
        if !possible {return false}
        var dp:[Set<Int>] = [Set<Int>](repeating:Set<Int>(),count:m + 1)
        dp[0].insert(0)
        for num in A
        {
            for i in stride(from:m,through:1,by:-1)
            {
                for a in dp[i - 1]
                {
                    dp[i].insert(a + num)
                }
            }
        }
        for i in 1...m
        {
            if sum * i % n == 0 && dp[i].contains(sum * i / n)
            {
                return true            
            }
        }
        return false
    }
}