725. Split Linked List in Parts

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, 
oot.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].

class Solution {
    public ListNode[] splitListToParts(ListNode root, int k) {
        ListNode[] res = new ListNode[k];
        ListNode cur = root;
        int le = 0;
        while(cur != null) {
            le++;
            cur = cur.next;
        }
        int rem = le % k, times = le / k, i = 0;
        cur = root;
        
        while(cur != null) {
            res[i++] = cur;
            for(int j = 1; j < times; j++) {
                cur = cur.next;
            }
            if(times > 0 && rem-- > 0) {
                cur = cur.next;
            }
            ListNode tmp = cur.next;
            cur.next = null;
            cur = tmp;
        }
        return res;
    }
}

把node as 均匀 as possible的分布在数组里

先得到长度，再得到一个数组index放多少个 times，得到放下times后还剩余多少rem，要把rem从前往后均匀的放。

进入循环，先满足放times个，然后如果rem有剩余就再往后挪一位，然后断开cur和后面的连接。直接cover了所有情况