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.
题目标签：Linked List
题目给了我们一个链表，还给了我们k，让我们把链表分成 k 个部分，使得每一个部分尽可能相等。
首先，我们要知道链表的长度 len，利用 len 和 k 来分组。
举例：
1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> 10
len = 10; k = 7 的话；
首先计算 base = len / k = 10 / 7 = 1；
然后计算 leftover = len % k = 10 % 7 = 3；
先把每一个小组的base 值填入
[1, 1, 1, 1, 1, 1, 1] 一共有7个小组，每个小组目前有1个node
然后把 leftover 分别加 1到每一个小组，从前到后，直到加完。
[2, 2, 2, 1, 1, 1, 1]
这样的话，一共有7个小组，前三个小组，每一组有2个 nodes，后面4个小组，每一组有1个node。
这里写成 array 方便理解，答案中只需要知道 base， 然后把 leftover 加上就可以。
所以根据上面的 array， 就可以把原链表分组为：
[1 -> 2] [3 -> 4] [5 -> 6] [7] [8] [9] [10]
Java Solution:
Runtime beats 59.09%
完成日期：12/07/2017
关键词：singly-linked list
关键点：计算出base 和 leftover
1 /**
2 * Definition for singly-linked list.
3 * public class ListNode {
4 * int val;
5 * ListNode next;
6 * ListNode(int x) { val = x; }
7 * }
8 */
9 class Solution
10 {
11 public ListNode[] splitListToParts(ListNode root, int k)
12 {
13 ListNode[] list = new ListNode[k];
14 int len = 0;
15 ListNode cursor = root;
16 int base = 0;
17 int leftover = 0;
18
19 // count the length
20 while(cursor != null)
21 {
22 cursor = cursor.next;
23 len++;
24 }
25
26 // calculate the base and leftover
27 base = len / k;
28 leftover = len % k;
29 cursor = root;
30
31 // iterate each group
32 for(int i=0; i<k; i++)
33 {
34 list[i] = cursor; // save this group head
35 ListNode tail = null;
36 int groupSize = base;
37
38 // set up correct group size
39 if(leftover > 0)
40 {
41 groupSize++;
42 leftover--;
43 }
44
45 // iterate this group nodes
46 for(int j=0; j<groupSize; j++)
47 {
48 if(j == groupSize - 1) // approach to the end of this group
49 tail = cursor;
50
51 cursor = cursor.next;
52 }
53
54 if(groupSize > 0) // link this group tail to null
55 tail.next = null;
56 }
57
58
59 return list;
60 }
61 }
参考资料：N/A
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