21 Merge Two Sorted Lists

---

title: Merge Two Sorted Lists
tags:
- merge-two-sorted-lists
- No.21
- simple
- merge
- loop-optimization
- list

---

Problem

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:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

Corner Cases

• empty list

Input: []
       []
return: []

Solutions

Merge Sort

Just the same as the merge part of merge sort. To avoid too much if-else in loop:

1. Take ListNode[] array and index to denote the target.
2. Initialize outside the loop.

The list update:

0.0. initialization with l1
[l1, p1, hd] -> x -> x
[l2, p2] -> x -> x -> x

0.1.
[l1, p1, hd] -> [np] -> x
[l2, p2] -> x -> x -> x

0.2.
[p1, np] -> x -> x -> x
[l2, p2] -> x -> x -> x
[hd, px]

1. before updating
[p1] -> x -> x -> x
[p2] -> x -> x -> x
[px]

2. choose p1
[p1] -> [np] -> x -> x
[p2] -> x -> x -> x
[px]

3.
[px] -> [p1] -> [np] -> x -> x
        [p2] -> x -> x -> x

4. 
x -> [p1, px] -> [np] -> x -> x
     [p2] -> x -> x -> x

5. 
[np] -> x -> x -> x
[p2] -> x -> x -> x
x -> [px]

6. 
[p1, np] -> x -> x -> x
[p2] -> x -> x -> x
x -> [px]

Running time is O(min(l_1, l_2)), since we directly copy the rest of l2 to target list hd when l1 comes to null first.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if      (l1 == null) { return l2; }
        else if (l2 == null) { return l1; }

        ListNode[] ps = new ListNode[] {l1, l2};
        int        id = (ps[0].val <= ps[1].val) ? 0 : 1;
        ListNode   hd = ps[id];
        ListNode   px = hd;
        ListNode   np = ps[id].next;
        ps[id]        = np;

        while (ps[0] != null && ps[1] != null) {
            id      = (ps[0].val <= ps[1].val) ? 0 : 1;
            np      = ps[id].next;
            px.next = ps[id];
            px      = px.next;
            px.next = null;
            ps[id]  = np;
        }
        // copy the residual list
        px.next = (ps[1] == null) ? ps[0] : ps[1];

        return hd;
    }
}