I tried LeetCode every day 21. Merge Two Sorted Lists (Python, Go)

Introduction

@Ishishow is running a free English word site E-tan.

I would like to work on letcode every day to improve my ability as a programmer and give my own way of solving it.

What is Leetcode

leetcode.com This is the practice of coding interviews for software developers. A total of more than 1,500 coding questions have been posted, and it seems that the same questions are often asked in actual interviews.

Introduction to Go language + algorithm I will solve it with Golang and Python to strengthen my brain. (Python is weak but experienced)

7th question (problem 21)

21. Merge Two Sorted Lists

--Problem content (Japanese translation)

Merges two sorted linked lists and returns them as a new ** sorted ** list. The new list must be created by joining the nodes of the first two lists together.

Example 1:

  Input: l1 = [1,2,4], l2 = [1,3,4]
  Output: [1,1,2,3,4,4]

Example 2:

  Input: l1 = [], l2 = []
  Output: []

Example 3:

  Input: l1 = [], l2 = [0]
  Output: [0]

Way of thinking

1. Create a new ListNode
2. Pass the created ListNode reference to a variable called res
3. Turn the loop, advance the smaller val of l1 and l2, and assign it to cur.
4. When either one goes to the end point, the loop ends, so substitute the rest.
5. return is res. Since it is 0 at first, it starts from the next.

--Answer code

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        cur = ListNode(0)
        res = cur
        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 or l2
        return res.next  

--I'll write it in Go too!

func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
	cur := &ListNode{}
	res := cur
	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 if l2 != nil {
		cur.Next = l2
	}

	return res.Next
}