725 Split Linked List in Parts

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725. Split Linked List in Parts

题目描述

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 ]

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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 [].
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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].

题目大意

给定数值k，把链条平均分配为k个链表，以数组形式返回。
记链表长度 % k为r，有以下规则：

  • 如果k > 链表长度，则返回k个元素的数组，不足的用nil补充。
  • 如果r > 0，则前r项多分配一个节点。
  • 如果r == 0，则链表平均分配节点数即可。

解题思路

先求得链表长度。
按照题目规则生成节点长度的数组，需要用nil补充的，标记为-1

代码

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/**
 * 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?] {
        
        //求长度
        func findlength(root: ListNode?) -> Int {
            var list = root
            var length = 0
            while list != nil {
                length += 1
                list = list?.next
            }
            return length
        }
        
        
        //求每个链表的长度
        func findListLength(length: Int) -> [Int] {
            let num = length / k
            let r = length % k
            let counts: [Int]
            if r > 0 && r < length {
                counts = Array(repeating: num + 1, count: r) + Array(repeating: num, count: k - r)
            }else if r == length {
                counts = Array(repeating: 1, count: length) + Array(repeating: -1, count: k - length)
            }else {
                counts = Array(repeating: num, count: k)
            }
            return counts
        }
        
        let counts = findListLength(length: findlength(root: root))
        var list = root
        var ans = [ListNode?]()
        for count in counts {
            if count == -1 {
                ans.append(nil)
                continue
            }
            var node = list
            for _ in 0..<count - 1 {
                node = node?.next
            }
            
            ans.append(list)
            list = node?.next
            node?.next = nil
        }
        return ans
    }
}
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