★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤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/10514474.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！★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
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].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 }
转载于:https://www.cnblogs.com/strengthen/p/10514474.html
