[Swift]LeetCode725. 分隔链表 | Split Linked List in Parts

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Given a (singly) linked list with head node root, write a function to split the linked list into k consecutive linked list "parts".

The length of each part should be as equal as possible: no two parts should have a size differing by more than 1. This may lead to some parts being null.

The parts should be in order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal parts occurring later.

Return a List of ListNode's representing the linked list parts that are formed.

Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ]

Example 1:

Input: 
root = [1, 2, 3], k = 5
Output: [[1],[2],[3],[],[]]
Explanation:
The input and each element of the output are ListNodes, not arrays.
For example, the input root has root.val = 1, root.next.val = 2, \root.next.next.val = 3, and root.next.next.next = null.
The first element output[0] has output[0].val = 1, output[0].next = null.
The last element output[4] is null, but it's string representation as a ListNode is [].

Example 2:

Input: 
root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
Explanation:
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.

Note:

• The length of root will be in the range [0, 1000].
• Each value of a node in the input will be an integer in the range [0, 999].
• k will be an integer in the range [1, 50].

---

给定一个头结点为 root 的链表, 编写一个函数以将链表分隔为 k 个连续的部分。

每部分的长度应该尽可能的相等: 任意两部分的长度差距不能超过 1，也就是说可能有些部分为 null。

这k个部分应该按照在链表中出现的顺序进行输出，并且排在前面的部分的长度应该大于或等于后面的长度。

返回一个符合上述规则的链表的列表。

举例： 1->2->3->4, k = 5 // 5 结果 [ [1], [2], [3], [4], null ]

示例 1：

输入: 
root = [1, 2, 3], k = 5
输出: [[1],[2],[3],[],[]]
解释:
输入输出各部分都应该是链表，而不是数组。
例如, 输入的结点 root 的 val= 1, root.next.val = 2, \root.next.next.val = 3, 且 root.next.next.next = null。
第一个输出 output[0] 是 output[0].val = 1, output[0].next = null。
最后一个元素 output[4] 为 null, 它代表了最后一个部分为空链表。

示例 2：

输入: 
root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
输出: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]]
解释:
输入被分成了几个连续的部分，并且每部分的长度相差不超过1.前面部分的长度大于等于后面部分的长度。

提示:

• root 的长度范围： [0, 1000].
• 输入的每个节点的大小范围：[0, 999].
• k 的取值范围： [1, 50].

---

Runtime: 20 ms

Memory Usage: 19.7 MB

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {
        var root = root
        var res:[ListNode?] = [ListNode?](repeating:nil,count:k)
        var len:Int = 0
        var t:ListNode? = root
        while(t != nil)
        {
            len += 1
            t = t!.next
        }
        var avg:Int = len / k
        var ext:Int = len % k
        var i:Int = 0
        while(i < k && root != nil)
        {
            res[i] = root
            var num:Int = (i < ext) ? 1 : 0
            for j in 1..<(avg + num )
            {
                root = root!.next
            }
            var t:ListNode? = root!.next
            root?.next = nil
            root = t   
            i += 1
        }
        return res
    }
}

---

20ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
    func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {
        var result  = Array<ListNode?>(repeating: nil, count: k)
        if nil == root {
            return result
        }
        var length = 0
        var temp : ListNode? = root
        while nil != temp {
            length += 1
            temp = temp?.next
        }
        if k >= length {
            temp = root
            var i = 0
            while nil != temp {
                let p = temp?.next
                temp?.next = nil
                result[i] = temp
                temp = p
                i += 1
            }
            return result
        }
        
        let average = length / k
        let mod =  length % k
        var currentHead : ListNode? = root
        for i in 0..<k {
            temp = currentHead
            let targetNodes = i < mod ? (average + 1) : average
            var j : Int = 1
            while j < targetNodes  {
                temp = temp?.next
                j += 1
            }
            result[i] = currentHead
            currentHead = temp?.next
            temp?.next = nil            
        }
        return result
    }
}

---

24ms

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.next = nil
 *     }
 * }
 */
class Solution {
func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {
    var current = root
    var count = 0
    
    while current != nil {
        current = current?.next
        count += 1
    }
    
    let width = count / k
    let rem = count % k
    
    var result: [ListNode?] = []
    
    current = root
    for i in 0..<k {
        let head = current
        let generate = width + (i < rem ? 1 : 0) - 1
        if generate < 0 {
            result.append(nil)
            continue
        }
        
        for _ in 0..<generate {
            if current != nil {
                current = current?.next
            }
        }
        if current != nil {
            let prev = current
            current = current?.next
            prev?.next = nil
        }
        result.append(head)
    }    
    return result
  }
}

---

19288kb

class Solution {
    func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {        
        if k == 0 {
            return [root]
        }        
        var current = root
        var length = 1
        while current?.next != nil {
            current = current?.next
            length += 1
        }        
        var elements = length / k
        var extras = length % k        
        current = root        
        var result = [ListNode?]()        
        if elements == 0 { // k >= length
            var counter = k            
            while current != nil {
                let node = ListNode(current!.val)
                result.append(node)
                current = current?.next
                counter -= 1
            }
            while counter > 0 {
                result.append(current)
                counter -= 1
            }
            return result
        }        
        var node = current
        while current != nil {            
            elements -= 1            
            if elements == 0 {                
                if extras > 0 {
                    extras -= 1
                    current = current?.next
                }                
                let next = current?.next                
                current?.next = nil                
                result.append(node)                
                current = next
                node = current                
                elements = length / k                
                continue
            }            
            current = current?.next
        }        
        return result
    }
}