链表问题（8）----合并

一、题目：合并两个有序的单链表

思路：时间复杂度O（M+N），空间复杂度O（1）

 简单来说就是在原来的链表上采用三个指针来操作两个链表。

若是合并两个无序链表成有序链表，先将两个链表用冒泡或快排等方法排序，再合并。

代码：

class Node:
    def __init__(self,value):
        self.val = value
        self.next = None
def mergeList(l1,l2):
    if not l1 or not l2:
        return l1 if l1 else l2
    cur1 , cur2 = l1 , l2
    head = cur1 if cur1.val < cur2.val else cur2
    cur = None
    while cur1 and cur2:
        tail = cur
        cur = cur1 if cur1.val < cur2.val else cur2
        cur1 = cur1.next if cur == cur1 else cur1
        cur2 = cur2.next if cur == cur2 else cur2
        if tail:
            tail.next = cur
    tail.next.next = cur1 if cur1 else cur2
    return head

    
    
head = Node(1)
p = head
for i in range(2,4):
    p.next = Node(i)
    p = p.next
head1 = Node(1)
p = head1
for i in range(2,5):
    p.next = Node(i)
    p = p.next
mergeList(head,head1)

二、题目：按照左右半区的方式重新组合单链表

思路：时间复杂度O（N），空间复杂度O（1）

简单来说就是将原链表拆分成左右半区，再合并。

代码：

class Node:
    def __init__(self,value):
        self.val = value
        self.next = None
#拆分链表
def relocate(head):
    if not head or not head.next:
        return head
    slow = head
    fast = head.next
    while fast.next and fast.next.next:
        slow = slow.next
        fast = fast.next.next
    right = slow.next
    slow.next = None
    mergeLR(head,right)
#合并链表
def mergeLR(left,right):
    if not left or not right:
        return left if left else right
    head = left
    cur =head
    while left:
        left = left.next
        cur.next = right 
        cur = cur.next
        right = right.next
        cur.next = left
        cur = cur.next
    return head
head = Node(1)
p = head
for i in range(2,9):
    p.next = Node(i)
    p = p.next
res = relocate(head)