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.

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:

题目大意：将一个链表分成k份，不能等分的时候，前面的链表可以比后面链表长一个节点。如果是不够k长，后面链表为空。

思路：1.先计算平均每份节点数，average=len/k;再查看是否存在不能均分情况，rem = len%k;rem大于0即不能均分。

2.再循环k次，每次将ret[i]指向当前节点。再向后移动average个节点，如果rem>0,再向后移动一个节点，同时rem减一；末尾指空。

代码：

class Solution {
    public ListNode[] splitListToParts(ListNode root, int k) {
        ListNode ret[] = new ListNode[k];
        if(root==null||root.next==null){
            ret[0]=root;
            return ret;
        }

        int len=0;
        ListNode head = root;
        while(head!=null){
            len++;
            head = head.next;
        }
        //System.out.println(len);
        int average = len/k;
        //System.out.println(average);
        int rem = 0;
        if(len>k)
        	rem = len%k;
        head = root;
        ListNode cur ;
        for(int i=0;i<k;i++){
            ret[i]=head;
            cur = head.next;
            for(int j=1;j<average;j++){
                head = head.next;
                if(head==null)
                    break;  
                cur = cur.next;
            }
            if(rem>0){
                head = head.next;
                cur = cur.next;           	
                rem--;
            }
            if(head!=null)
            	head.next=null;
            head = cur;
            if(head==null)
                break;            
            cur = cur.next;
        }
          
        return ret;
    }
}