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86 lines
2.4 KiB
Python
86 lines
2.4 KiB
Python
#!/usr/bin/env python
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# coding=utf-8
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#######################################################################################
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# Leetcode 148 排序链表
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#
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# 在 O(n log n) 时间复杂度和常数级空间复杂度下，对链表进行排序。
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#
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# 示例 1:
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# 输入: 4->2->1->3
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# 输出: 1->2->3->4
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#
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# 示例 2:
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# 输入: -1->5->3->4->0
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# 输出: -1->0->3->4->5
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#######################################################################################
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class ListNode:
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def __init__(self, x):
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self.val = x
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self.next = None
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def __repr__(self):
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if self:
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return "{}->{}".format(self.val, repr(self.next))
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class Solution:
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def sortList(self, head):
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"""
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:type head: ListNode
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:rtype ListNode
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(knowledge)
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思路：
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1. 使用递归方式实现链表的归并排序算法；
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2. 对链表进行分治的时候，可以使用快慢指针（快指针一次走两步，慢指针一次走一步，同时从head出发，当快指针走到链表尾的时候，慢指针就走到中间位置）；
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"""
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# 链表为空或者只有一个元素，直接返回
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if not head or not head.next:
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return head
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# low -> 右子链的头部
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# pre -> 左子链最后一个节点（未断开之前，指向low）
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low, fast, pre = head, head, None
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while fast and fast.next:
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low, fast, pre = low.next, fast.next.next, low
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# 断开左右子链之间的链接
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pre.next = None
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# 对左右子链分别递归的进行归并排序
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head = self.sortList(head)
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low = self.sortList(low)
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# 对排好序的两部分进行归并
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if not low:
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return head
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result, cur = self, None
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while head and low:
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if head.val <= low.val:
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result.next, head = head, head.next
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else:
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result.next, low = low, low.next
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result = result.next
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if not head:
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result.next = low
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elif not low:
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result.next = head
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else:
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result.next = None
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return self.next
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if __name__ == '__main__':
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solution = Solution()
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h1 = ListNode(4)
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h1.next = ListNode(2)
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h1.next.next = ListNode(1)
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h1.next.next.next = ListNode(3)
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print(solution.sortList(h1), "= 1->2->3->4")
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