0021._merge_two_sorted_lists.md

Links:

1. https://leetcode.com/problems/merge-two-sorted-lists/
2. https://leetcode-cn.com/problems/merge-two-sorted-lists/

Tags

1. 链表

Solution 1 递归

每次选择两个链表头部较小的一个节点，然后剩下的节点和另外一个链表的元素继续merge.

子问题为去掉当前最小的元素，剩下的元素继续merge。

# pseudo code
if list1[0] < list2[0]:
    list1[0] + merge(list1[1:], list2)
else:
    list2[0] + merge(list1, list2[1:])

---

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        if l1 is None:
            return l2
            
        if l2 is None:
            return l1

        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2

---

Go

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */

func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	}

	if l2 == nil {
		return l1
	}

	if l1.Val < l2.Val {
		l1.Next = mergeTwoLists(l1.Next, l2)
		return l1
	}

	l2.Next = mergeTwoLists(l2.Next, l1)
	return l2
}

---

Go

func mergeTwoLists(l1, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	}
	if l2 == nil {
		return l1
	}
	n := &ListNode{}
	if l1.Val < l2.Val {
		n.Val = l1.Val
		n.Next = mergeTwoLists1(l1.Next, l2)
	} else {
		n.Val = l2.Val
		n.Next = mergeTwoLists1(l1, l2.Next)
	}
	return n
}

Solution2 递归

l1 始终保存最小的元素的链表

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        if l1 and l2:
            if l1.val > l2.val:
                 l1, l2 = l2, l1
            l1.next = self.mergeTwoLists(l1.next, l2)
        return l1 or l2

Solution3 迭代（循环）

用一个指针保存当前的位置，每一次选出一个最小的元素，插入到当前的位置。

其实是用迭代改写Solution1的递归，思路一样。

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

class Solution:
    def mergeTwoLists(self, l1, l2):
        if l1 is None:
            return l2

        if l2 is None:
            return l1

        dummy = cur = ListNode(None)

        while l1 and l2:
            if l1.val < l2.val:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next

            cur = cur.next

        cur.next = l1 if l1 else l2
        return dummy.next

---

Go

func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
	if l1 == nil {
		return l2
	}

	if l2 == nil {
		return l1
	}

	dummy := &ListNode{}
	cur := dummy

	for l1 != nil && l2 != nil {
		if l1.Val < l2.Val {
			cur.Next = l1
			l1 = l1.Next
		} else {
			cur.Next = l2
			l2 = l2.Next
		}

		cur = cur.Next
	}

	if l1 != nil {
		cur.Next = l1
	} else {
		cur.Next = l2
	}

	return dummy.Next
}