Skip to content

725. Split Linked List in Parts

DifficultyMedium

Description

Given the head of a singly linked list and an integer k, 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 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

Solutions

Solution 1: Simulation

Thinking

Split a list into \(k\) parts as evenly as possible, giving leftover nodes to the earlier parts. \(n\le 1000\), so count then cut.

Each part gets \(\lfloor n/k\rfloor\) nodes, and the first \(n\bmod k\) parts get one extra. Unused slots stay null.

A second walk advances the computed length, severs \(\textit{next}\), and records the next head. Two passes in all.

First, we traverse the linked list to obtain its length \(n\), and then we calculate the average length \(\textit{cnt} = \lfloor \frac{n}{k} \rfloor\) and the remainder \(\textit{mod} = n \bmod k\). For the first \(\textit{mod}\) parts, each part has a length of \(\textit{cnt} + 1\), while the lengths of the remaining parts are \(\textit{cnt}\).

Next, we just need to traverse the linked list and split it into \(k\) parts.

The time complexity is \(O(n)\), and the space complexity is \(O(k)\). Here, \(n\) is the length of the linked list.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
# Definition for singly-linked list.
# 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]]:
        n = 0
        cur = head
        while cur:
            n += 1
            cur = cur.next
        cnt, mod = divmod(n, k)
        ans = [None] * k
        cur = head
        for i in range(k):
            if cur is None:
                break
            ans[i] = cur
            m = cnt + int(i < mod)
            for _ in range(1, m):
                cur = cur.next
            nxt = cur.next
            cur.next = None
            cur = nxt
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode[] splitListToParts(ListNode head, int k) {
        int n = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            ++n;
        }
        int cnt = n / k, mod = n % k;
        ListNode[] ans = new ListNode[k];
        ListNode cur = head;
        for (int i = 0; i < k && cur != null; ++i) {
            ans[i] = cur;
            int m = cnt + (i < mod ? 1 : 0);
            for (int j = 1; j < m; ++j) {
                cur = cur.next;
            }
            ListNode nxt = cur.next;
            cur.next = null;
            cur = nxt;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    vector<ListNode*> splitListToParts(ListNode* head, int k) {
        int n = 0;
        for (ListNode* cur = head; cur != nullptr; cur = cur->next) {
            ++n;
        }
        int cnt = n / k, mod = n % k;
        vector<ListNode*> ans(k, nullptr);
        ListNode* cur = head;
        for (int i = 0; i < k && cur != nullptr; ++i) {
            ans[i] = cur;
            int m = cnt + (i < mod ? 1 : 0);
            for (int j = 1; j < m; ++j) {
                cur = cur->next;
            }
            ListNode* nxt = cur->next;
            cur->next = nullptr;
            cur = nxt;
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func splitListToParts(head *ListNode, k int) []*ListNode {
    n := 0
    for cur := head; cur != nil; cur = cur.Next {
        n++
    }

    cnt := n / k
    mod := n % k
    ans := make([]*ListNode, k)
    cur := head

    for i := 0; i < k && cur != nil; i++ {
        ans[i] = cur
        m := cnt
        if i < mod {
            m++
        }
        for j := 1; j < m; j++ {
            cur = cur.Next
        }
        next := cur.Next
        cur.Next = nil
        cur = next
    }

    return ans
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function splitListToParts(head: ListNode | null, k: number): Array<ListNode | null> {
    let n = 0;
    for (let cur = head; cur !== null; cur = cur.next) {
        n++;
    }
    const cnt = (n / k) | 0;
    const mod = n % k;
    const ans: Array<ListNode | null> = Array(k).fill(null);
    let cur = head;
    for (let i = 0; i < k && cur !== null; i++) {
        ans[i] = cur;
        let m = cnt + (i < mod ? 1 : 0);
        for (let j = 1; j < m; j++) {
            cur = cur.next!;
        }
        let next = cur.next;
        cur.next = null;
        cur = next;
    }
    return ans;
}

Comments