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
rootwill be in the range[0, 1000]. - Each value of a node in the input will be an integer in the range
[0, 999]. kwill 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
1 /**
2 * Definition for singly-linked list.
3 * public class ListNode {
4 * public var val: Int
5 * public var next: ListNode?
6 * public init(_ val: Int) {
7 * self.val = val
8 * self.next = nil
9 * }
10 * }
11 */
12 class Solution {
13 func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {
14 var root = root
15 var res:[ListNode?] = [ListNode?](repeating:nil,count:k)
16 var len:Int = 0
17 var t:ListNode? = root
18 while(t != nil)
19 {
20 len += 1
21 t = t!.next
22 }
23 var avg:Int = len / k
24 var ext:Int = len % k
25 var i:Int = 0
26 while(i < k && root != nil)
27 {
28 res[i] = root
29 var num:Int = (i < ext) ? 1 : 0
30 for j in 1..<(avg + num )
31 {
32 root = root!.next
33 }
34 var t:ListNode? = root!.next
35 root?.next = nil
36 root = t
37 i += 1
38 }
39 return res
40 }
41 }
20ms
1 /**
2 * Definition for singly-linked list.
3 * public class ListNode {
4 * public var val: Int
5 * public var next: ListNode?
6 * public init(_ val: Int) {
7 * self.val = val
8 * self.next = nil
9 * }
10 * }
11 */
12 class Solution {
13 func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {
14 var result = Array<ListNode?>(repeating: nil, count: k)
15 if nil == root {
16 return result
17 }
18 var length = 0
19 var temp : ListNode? = root
20 while nil != temp {
21 length += 1
22 temp = temp?.next
23 }
24 if k >= length {
25 temp = root
26 var i = 0
27 while nil != temp {
28 let p = temp?.next
29 temp?.next = nil
30 result[i] = temp
31 temp = p
32 i += 1
33 }
34 return result
35 }
36
37 let average = length / k
38 let mod = length % k
39 var currentHead : ListNode? = root
40 for i in 0..<k {
41 temp = currentHead
42 let targetNodes = i < mod ? (average + 1) : average
43 var j : Int = 1
44 while j < targetNodes {
45 temp = temp?.next
46 j += 1
47 }
48 result[i] = currentHead
49 currentHead = temp?.next
50 temp?.next = nil
51 }
52 return result
53 }
54 }
24ms
1 /**
2 * Definition for singly-linked list.
3 * public class ListNode {
4 * public var val: Int
5 * public var next: ListNode?
6 * public init(_ val: Int) {
7 * self.val = val
8 * self.next = nil
9 * }
10 * }
11 */
12 class Solution {
13 func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {
14 var current = root
15 var count = 0
16
17 while current != nil {
18 current = current?.next
19 count += 1
20 }
21
22 let width = count / k
23 let rem = count % k
24
25 var result: [ListNode?] = []
26
27 current = root
28 for i in 0..<k {
29 let head = current
30 let generate = width + (i < rem ? 1 : 0) - 1
31 if generate < 0 {
32 result.append(nil)
33 continue
34 }
35
36 for _ in 0..<generate {
37 if current != nil {
38 current = current?.next
39 }
40 }
41 if current != nil {
42 let prev = current
43 current = current?.next
44 prev?.next = nil
45 }
46 result.append(head)
47 }
48 return result
49 }
50 }
19288kb
1 class Solution {
2 func splitListToParts(_ root: ListNode?, _ k: Int) -> [ListNode?] {
3 if k == 0 {
4 return [root]
5 }
6 var current = root
7 var length = 1
8 while current?.next != nil {
9 current = current?.next
10 length += 1
11 }
12 var elements = length / k
13 var extras = length % k
14 current = root
15 var result = [ListNode?]()
16 if elements == 0 { // k >= length
17 var counter = k
18 while current != nil {
19 let node = ListNode(current!.val)
20 result.append(node)
21 current = current?.next
22 counter -= 1
23 }
24 while counter > 0 {
25 result.append(current)
26 counter -= 1
27 }
28 return result
29 }
30 var node = current
31 while current != nil {
32 elements -= 1
33 if elements == 0 {
34 if extras > 0 {
35 extras -= 1
36 current = current?.next
37 }
38 let next = current?.next
39 current?.next = nil
40 result.append(node)
41 current = next
42 node = current
43 elements = length / k
44 continue
45 }
46 current = current?.next
47 }
48 return result
49 }
50 }
原文地址：https://www.cnblogs.com/strengthen/p/10514474.html
时间： 2024-07-29 12:08:27