[Swift]LeetCode473. 火柴拼正方形 | Matchsticks to Square

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/10347635.html 
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎，请直接点击！！！

进入博主App Store主页，下载使用各个作品！！！

注：博主将坚持每月上线一个新app！！！

Remember the story of Little Match Girl? By now, you know exactly what matchsticks the little match girl has, please find out a way you can make one square by using up all those matchsticks. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Your input will be several matchsticks the girl has, represented with their stick length. Your output will either be true or false, to represent whether you could make one square using all the matchsticks the little match girl has.

Example 1:

Input: [1,1,2,2,2]
Output: true

Explanation: You can form a square with length 2, one side of the square came two sticks with length 1. 

Example 2:

Input: [3,3,3,3,4]
Output: false

Explanation: You cannot find a way to form a square with all the matchsticks. 

Note:

1. The length sum of the given matchsticks is in the range of 0 to 10^9.
2. The length of the given matchstick array will not exceed 15.

---

还记得童话《卖火柴的小女孩》吗？现在，你知道小女孩有多少根火柴，请找出一种能使用所有火柴拼成一个正方形的方法。不能折断火柴，可以把火柴连接起来，并且每根火柴都要用到。

输入为小女孩拥有火柴的数目，每根火柴用其长度表示。输出即为是否能用所有的火柴拼成正方形。

示例 1:

输入: [1,1,2,2,2]
输出: true

解释: 能拼成一个边长为2的正方形，每边两根火柴。

示例 2:

输入: [3,3,3,3,4]
输出: false

解释: 不能用所有火柴拼成一个正方形。

注意:

1. 给定的火柴长度和在 0 到 10^9之间。
2. 火柴数组的长度不超过15。

---

Runtime: 836 ms

Memory Usage: 3.9 MB

class Solution {
    func makesquare(_ nums: [Int]) -> Bool {
        if nums.isEmpty || nums.count < 4 {return false}
        var sum:Int = nums.reduce(0, +)
        if sum % 4 != 0 {return false}
        var n:Int = nums.count
        var all:Int = (1 << n) - 1
        var target:Int = sum / 4
        var masks:[Int] = [Int]()
        var validHalf:[Bool] = [Bool](repeating:false,count:1 << n)
        for i in 0...all
        {
            var curSum:Int = 0
            for j in 0...15
            {
                if ((i >> j) & 1) == 1
                {
                    curSum += nums[j]
                }
            }
            if curSum == target
            {
                for mask in masks
                {
                    if (mask & i) != 0 {continue}
                    var half:Int = mask | i
                    validHalf[half] = true
                    if validHalf[all ^ half] {return true}
                }
                masks.append(i)
            }
        }
        return false
    }
}

---

 3911 kb

class Solution {
    func makesquare(_ nums: [Int]) -> Bool {
        guard  nums.count >= 4 else { return false }
        let sum = nums.reduce(0, { $0 + $1 })
        if sum % 4 != 0 {
            return false
        }
        var sides: [Int] = Array(repeating: 0, count: 4)
        return dfs(nums.sorted(by: >), 0, &sides, sum / 4)
    }
    
    func dfs(_ nums: [Int], _ index: Int, _ sides: inout [Int], _ target: Int) -> Bool {
        if index == nums.count {
            return sides[0] == sides[1] && sides[0] == sides[2] && sides[0] == sides[3] && sides[0] == target
        }
        
        for i in 0 ..< sides.count {
            if sides[i] + nums[index] > target {
                continue
            }
            sides[i] += nums[index]
            if dfs(nums, index + 1, &sides, target) {
                return true
            }
            sides[i] -= nums[index]
        }
        return false
    }
}