【python3】leetcode 725. Split Linked List in Parts (Medium)_split linked list in parts(medium)-优快云博客

本文链接：https://blog.youkuaiyun.com/maotianyi941005/article/details/86302247

本文介绍了如何解决LeetCode 725题——将链表拆分成连续的部分。通过遍历链表获取长度，计算平均长度和余数，确保各部分大小相差不超过1，并保持原有的顺序。实现的时间复杂度为O(n+k)，空间复杂度为O(k)。

725. Split Linked List in Parts (Medium)

Given a (singly) linked list with head node root, write a function to split the linked list into kconsecutive 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.
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].

1 遍历得到len

2 计算每组平均的长度 avglen = len / k

3 余数remainder，对前remainder组，每组长度多加1个

time:O(n+k) , space:O(k)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def splitListToParts(self, root, k):
        """
        :type root: ListNode
        :type k: int
        :rtype: List[ListNode]
        """
        n = 0 #len of link list
        r = root   
        parts = []
        while(r):
            n += 1
            r = r.next
        avglen,remainder = divmod(n , k)
        
        while(root):
            subpart = None
            for i in range(avglen):
                if not subpart: subpart = root#only head node of this subpart
                pre = root
                root = root.next
            if remainder >0 :
                if not subpart: subpart = root
                pre = root
                root = root.next
                remainder -= 1
            parts.append(subpart)
            if parts:pre.next = None
        if n < k:
            parts.extend([[]]*(k - n))  
        return parts