725. Split Linked List in Parts

https://leetcode.com/problems/split-linked-list-in-parts/

k>=length, 每个部分有且仅有一个，剩下k-length个部分都是空

Input: 
root = [1, 2, 3], k = 5
Output: [[1],[2],[3],[],[]]

k<length，每个部分至少有length/k个，那么剩下的length%k个就要给每个部分匀一个

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]]

最终有：

前length%k个部分，size为（length/k）+1

剩下的部分size就是length/k

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode[] splitListToParts(ListNode root, int k) {
        ListNode[] result = new ListNode[k];
        int length = 0;
        for (ListNode cursor = root; cursor != null; cursor = cursor.next) {
            length++;
        }
        int perPartNum = length / k;
        int remainNum = length % k;
        ListNode head = root;
        ListNode perPartTail = null;
        for(int i = 0; i < k; i++, remainNum--){
            result[i] = head;
            for(int j = 0; j<perPartNum+(remainNum> 0 ? 1 : 0); j++){
                perPartTail = head;
                head = head.next;
            }
            if(perPartTail != null){
                perPartTail.next = null;
            }
        }
        return result;
    }
}

remainNum就是需要匀出去的部分，最开始的几个在分配的时候就会多一个