Formatted question description: https://leetcode.ca/all/725.html

Medium

Description

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].

Solution

First obtain the length of the linked list, and calculate each part’s length after splitting the list into k parts.

Then for each part, determine the head of each part and put the nodes in the part with the part’s size.

• /**

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].

*/

public static void main(String[] args) {
Solution s = out.new Solution();

ListNode root = new ListNode(1);
root.next = new ListNode(2);
root.next.next = new ListNode(3);
root.next.next.next = new ListNode(4);
root.next.next.next.next = new ListNode(5);
root.next.next.next.next.next = new ListNode(6);
root.next.next.next.next.next.next = new ListNode(7);

System.out.println(s.splitListToParts(root, 3));
}

/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/

/*
corner case:
input: [], 3
expected: [[],[],[]]
*/
class Solution {
public ListNode[] splitListToParts(ListNode root, int k) {

if (root == null) {
return null;
}

//            if (root.next == null) {
//                return new ListNode[]{root};
//            }

int count = 0;
ListNode current =  root;
while (current != null) {
current = current.next;
count++;
}

int oneMoreCount = count % k;
int eachSize = count / k;
ListNode[] result = new ListNode[k];

current =  root;
for (int i = 0; i < k; i++) {

// set head of this part
result[i] = current;

// find all nodes of this part
int partRequired = eachSize + (oneMoreCount > 0 ? 1 : 0);
int partCount = 0;
ListNode partPrev = new ListNode(0);
ListNode partCurrent = current;
while (partCount < partRequired && partCurrent != null) {
partPrev = partCurrent;
partCurrent = partCurrent.next;
partCount++;
}
// now partCurrent is at head of next part
current = partCurrent;

partPrev.next = null;
oneMoreCount--;
}

return result;
}
}

}

############

/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode[] splitListToParts(ListNode root, int k) {
int n = 0;
ListNode cur = root;
while (cur != null) {
++n;
cur = cur.next;
}
// width 表示每一部分至少含有的结点个数
// remainder 表示前 remainder 部分，每一部分多出一个数
int width = n / k, remainder = n % k;
ListNode[] res = new ListNode[k];
cur = root;
for (int i = 0; i < k; ++i) {
for (int j = 0; j < width + ((i < remainder) ? 1 : 0) - 1; ++j) {
if (cur != null) {
cur = cur.next;
}
}
if (cur != null) {
ListNode t = cur.next;
cur.next = null;
cur = t;
}
}
return res;
}
}

• // OJ: https://leetcode.com/problems/split-linked-list-in-parts/
// Time: O(N)
// Space: O(1)
class Solution {
int ans = 0;
return ans;
}
public:
vector<ListNode*> splitListToParts(ListNode* head, int k) {
int len = getLength(head), d = len / k, r = len % k;
vector<ListNode*> ans;
for (; k; --k) {
int cnt = d + (r > 0);
if (r > 0) --r;
}
return ans;
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
def splitListToParts(self, root: ListNode, k: int) -> List[ListNode]:
n, cur = 0, root
while cur:
n += 1
cur = cur.next
cur = root
width, remainder = divmod(n, k)
res = [None for _ in range(k)]
for i in range(k):
for j in range(width + (i < remainder) - 1):
if cur:
cur = cur.next
if cur:
cur.next, cur = None, cur.next
return res

############

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
def splitListToParts(self, root, k):
"""
:type root: ListNode
:type k: int
:rtype: List[ListNode]
"""
nodes = []
counts = 0
each = root
while each:
counts += 1
each = each.next
num = counts / k
rem = counts % k
for i in range(k):
for j in range(num):
node = ListNode(root.val)
each.next = node
each = each.next
root = root.next
if rem and root:
rmnode = ListNode(root.val)
each.next = rmnode
if root:
root = root.next
rem -= 1
return nodes

• /**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode[] splitListToParts(ListNode root, int k) {
if (root == null)
return new ListNode[k];
if (k == 1)
return new ListNode[]{root};
int length = 0;
ListNode counter = root;
while (counter != null) {
counter = counter.next;
length++;
}
int[] sizes = new int[k];
int size = length / k;
for (int i = 0; i < k; i++)
sizes[i] = size;
int remainder = length % k;
for (int i = 0; i < remainder; i++)
sizes[i]++;
ListNode[] array = new ListNode[k];
for (int i = 0; i < k; i++) {
int curSize = sizes[i];
for (int j = 1; j <= curSize; j++) {
if (j == curSize) {
temp.next = null;
} else
temp = temp.next;
}
}
return array;
}
}

############

/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode[] splitListToParts(ListNode root, int k) {
int n = 0;
ListNode cur = root;
while (cur != null) {
++n;
cur = cur.next;
}
// width 表示每一部分至少含有的结点个数
// remainder 表示前 remainder 部分，每一部分多出一个数
int width = n / k, remainder = n % k;
ListNode[] res = new ListNode[k];
cur = root;
for (int i = 0; i < k; ++i) {
for (int j = 0; j < width + ((i < remainder) ? 1 : 0) - 1; ++j) {
if (cur != null) {
cur = cur.next;
}
}
if (cur != null) {
ListNode t = cur.next;
cur.next = null;
cur = t;
}
}
return res;
}
}

• // OJ: https://leetcode.com/problems/split-linked-list-in-parts/
// Time: O(N)
// Space: O(1)
class Solution {
int ans = 0;
return ans;
}
public:
vector<ListNode*> splitListToParts(ListNode* head, int k) {
int len = getLength(head), d = len / k, r = len % k;
vector<ListNode*> ans;
for (; k; --k) {
int cnt = d + (r > 0);
if (r > 0) --r;
}
return ans;
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
def splitListToParts(self, root: ListNode, k: int) -> List[ListNode]:
n, cur = 0, root
while cur:
n += 1
cur = cur.next
cur = root
width, remainder = divmod(n, k)
res = [None for _ in range(k)]
for i in range(k):
for j in range(width + (i < remainder) - 1):
if cur:
cur = cur.next
if cur:
cur.next, cur = None, cur.next
return res

############

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
def splitListToParts(self, root, k):
"""
:type root: ListNode
:type k: int
:rtype: List[ListNode]
"""
nodes = []
counts = 0
each = root
while each:
counts += 1
each = each.next
num = counts / k
rem = counts % k
for i in range(k):
for j in range(num):
node = ListNode(root.val)
each.next = node
each = each.next
root = root.next
if rem and root:
rmnode = ListNode(root.val)
each.next = rmnode
if root:
root = root.next
rem -= 1
return nodes