题目浅析

  • 想查看原题可以点击题目链接。

  • 简单地说，就是给一个链表和数字 k，现在要把这个链表分成 k 份，要求每份的节点数尽量平均，前面的节点数一定不小于后面的，相差不超过 1.

思路分享

代码解答（强烈建议自行解答后再看）

  • 参考题解
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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def splitListToParts(self, head: Optional[ListNode], k: int) -> List[Optional[ListNode]]:
        p = head
        n = 0
        while p != None:
            p = p.next
            n += 1
        base_part_size, need_plus = n//k, n%k

        ans = [None]*k
        i = 0
        cur = head
        
        while i < k and cur:
            part_size = base_part_size + (1 if i < need_plus else 0)
            ans[i] = cur
            pre = None
            while cur and part_size > 0:
                pre = cur
                cur = cur.next
                part_size -= 1
            if pre:
                pre.next = None
            i += 1
            
        return ans