“This is the 11th day of my participation in the Gwen Challenge in November. See details: The Last Gwen Challenge in 2021”

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Preface 🌧 ️

Algorithms are unfamiliar and familiar to the front-end people, and often we don’t value them as much as the back-end engineers do. But in fact, algorithms have an unshakable position for every programmer.

Because the development process is to convert the actual problem into the computer can recognize the instructions, that is, “data structure” said, “design the data structure, in the application of algorithms on the line”.

The quality of writing instructions will directly affect the performance of the program, and instructions are composed of data structure and algorithm, so the design of data structure and algorithm basically determines the quality of the final program.

In addition, when reading the source code, the lack of understanding of algorithms and data structures makes it difficult to understand why the author wrote the way he did.

In today’s environment, algorithms have become an indispensable skill for front-end engineers. If we want to move beyond being application engineers writing business code, we need to understand algorithms and data structures.

Of course, learning is also focused, as the front end we do not need to fully grasp the algorithm like back-end development, some of the more partial, not practical type and solution method, as long as a little understanding.

The title 🦀

21. Merge two ordered lists

Difficulty is simple

Merges two ascending lists into a new ascending list and returns. A new list is formed by concatenating all the nodes of a given two lists.

Example 1:

Input: l1 = [1,2,4], l2 = [1,3,4]Copy the code

Example 2:

Input: L1 = [], L2 = [] output: []Copy the code

Example 3:

Input: L1 = [], L2 = [0] output: [0]Copy the code

Tip:

  • The number of nodes in two linked lists ranges from[0, 50]
  • -100 <= Node.val <= 100
  • l1 和 l2All pressNondecreasing orderarrangement

🌵

  • Similar to merging two ordered arrays in merge sort
  • Replacing an array with a linked list solves this problem

How to solve the problem 🐂

  • Create a new linked list as the result.
  • Use a pointer to traverse two ordered lists, and compare the current nodes of the two lists. The smaller one is connected to the new list, and move the pointer one step back.
  • List traversal ends, returns a new list

Source 🔥

Depth-first ergodic solution

/** * Definition for singly-linked list. * function ListNode(val, next) { * this.val = (val===undefined ? 0 : val) * this.next = (next===undefined ? null : next) * } */
/ * * *@param {ListNode} l1
 * @param {ListNode} l2
 * @return {ListNode}* /
var mergeTwoLists = function(l1, l2) {
    const res=new ListNode(0);
    let p=res;
    let p1=l1;
    let p2=l2;

    while(p1 && p2){
        if(p1.val<p2.val){
          p.next=p1;
          p1=p1.next;
        }else{
            p.next=p2;
            p2=p2.next
        }
        p=p.next
    }
    if(p1){
        p.next=p1
    }
    if(p2){
        p.next=p2
    }
    return res.next

};
         
Copy the code

Time complexity: sum of the lengths of O(n) list one and list two

Space complexity :O(1)

Conclusion 🌞

So the “LeetCode” 21- merging two ordered linked lists ⚡️ is over. There is no shortcut to algorithm, so we can only write and practice more and summarize more. The purpose of this article is actually very simple, which is to urge myself to complete algorithm practice and summarize and output. I hope everyone can like my essay, and I also hope to know more like-minded friends through the article. If you also like to toss, welcome to add my friend, sand sculpture together, together progress.

Making 🤖 : sudongyu

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