合并两个已排序的链表

Merge Two Sorted Lists

• 合并两个已排序的链表，新链表中的每个节点必须是来自输入的两个链表的节点（即不能构造新的节点），返回新链表的头部。
• Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

example 1

input: 1->2->4, 3->8output: 1->2->3->4->8

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思路

1. head指向输入两个链表中头节点较小值，作为新链表的头部
2. tail指向新链表表尾，初始状态head = tail
3. a扫描l1，b扫描l2，比较a和b节点内值的大小，将较小的加入tail之后，a和b中较小的向后移动一个节点，较大的不动，tail向后移动一个节点保证任意时候指向都是新链表尾部
4. l1和l2其中一个已经遍历完，若另一个还有元素，添加到tail之后

代码

# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def mergeTwoLists(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        if None in (l1, l2):            return l1 or l2        head = tail = l1 if l1.val <= l2.val else l2        a = l1 if l1.val > l2.val else l1.next        b = l2 if l1.val <= l2.val else l2.next        while a and b:            if a.val <= b.val:                tail.next = a                tail, a = tail.next, a.next            else:                tail.next = b                tail, b = tail.next, b.next        tail.next = a or b        return head

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