# 725. Split Linked List in Parts

### 題目敘述

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

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

Return an array of the k parts.

Example 1:

Input: head = [1,2,3], k = 5
Output: [[1],[2],[3],[],[]]
Explanation:
The first element output[0] has output[0].val = 1, output[0].next = null.
The last element output[4] is null, but its string representation as a ListNode is [].

Example 2:

Input: head = [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.

Constraints:

• The number of nodes in the list is in the range [0, 1000].
• 0 <= Node.val <= 1000
• 1 <= k <= 50

Hint 1:

If there are N nodes in the list, and k parts, then every part has N/k elements, except the first N%k parts have an extra one.

### 解法解析

Time complexity：O(N + k). Space complexity：O(k)

#### 程式範例

##### Python
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def splitListToParts(self, head: Optional[ListNode], k: int) -> List[Optional[ListNode]]:
for N in range(1001):
if not cur:
break
cur = cur.next
width, remainder = divmod(N, k)

ans = []
for i in range(k):
tmp = cur
for _ in range(width + (i < remainder) - 1):
if cur:
cur = cur.next
if cur:
cur.next, cur = None, cur.next
ans.append(tmp)
return ans
##### JavaScript
/**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {number} k
* @return {ListNode[]}
*/
var splitListToParts = function (head, k) {
let len = 0;
while (cur) {
++len;
cur = cur.next;
}

let remainder = 0;
let width = Math.floor(len / k);
if (width === 0) width = 1;
else remainder = len % k;

let result = [];
for (let i = 0; i < k; i++) {
const tmp = cur;

let j = 0;
while (j < width + (i < remainder) - 1) {
if (cur) {
cur = cur.next;
}
j++;
}

if (cur) {
const temp = cur.next;
cur.next = null;
cur = temp;
}

result.push(tmp);
}
return result;
};
##### Go
/**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
func splitListToParts(head *ListNode, k int) []*ListNode {
len := 0
for cur := head; cur != nil; cur = cur.Next {
len++
}

width, remainder := len/k, len%k

ans := make([]*ListNode, k)
for i := 0; i < k && cur != nil; i++ {
ans[i] = cur
var rc int
if remainder > 0 {
rc = 1
}

for j := 1; j < width+rc; j++ {
cur = cur.Next
}

remainder--
t := cur.Next
cur.Next = nil
cur = t
}

return ans
}
##### Swift
/**
* public class ListNode {
*     public var val: Int
*     public var next: ListNode?
*     public init() { self.val = 0; self.next = nil; }
*     public init(_ val: Int) { self.val = val; self.next = nil; }
*     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
* }
*/
class Solution {
func splitListToParts(_ head: ListNode?, _ k: Int) -> [ListNode?] {
var ret = [ListNode?]()
var N = 0
while node != nil {
node = node!.next
N += 1
}

var r = N % k, q = N / k

for i in 0..<k {
ret.append(cur)
var size = q + (i < r ? 1 : 0)
var j = 1
while j < size {
cur = cur?.next
j += 1
}
let next = cur?.next
cur?.next = nil
cur = next
}

return ret
}
}
##### Kotlin
/**
* Example: var li = ListNode(5) var v = li.`val` Definition for singly-linked list. class
* ListNode(var `val`: Int) {
*
• var next: ListNode? = null
• ```
• } */ class Solution { fun splitListToParts(head: ListNode?, k: Int): Array {

val splitSize = listSize / k
val plusOne = listSize % k

val result = Array<ListNode?>(k) { null }

for (i in 1..k) {
result[i - 1] = itr
itr = run(itr, splitSize - 1 + if (plusOne >= i) 1 else 0)
val next = itr?.next
itr?.next = null
itr = next
}

return result

}

private fun size(head: ListNode?): Int {

var count = 0

while (itr != null) {
itr = itr.next
count++
}

return count

}

private fun run(head: ListNode?, num: Int): ListNode? {