21. Merge Two Sorted Lists
use std::borrow::Borrow; /** 21. Merge Two Sorted Lists https://leetcode.com/problems/merge-two-sorted-lists/ You are given the heads of two sorted linked lists list1 and list2. Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists. Return the head of the merged linked list. Example 1: Input: list1 = [1,2,4], list2 = [1,3,4] Output: [1,1,2,3,4,4] Example 2: Input: list1 = [], list2 = [] Output: [] Example 3: Input: list1 = [], list2 = [0] Output: [0] Constraints: 1. The number of nodes in both lists is in the range [0, 50]. 2. -100 <= Node.val <= 100 3. Both list1 and list2 are sorted in non-decreasing order. */ // Definition for singly-linked list. #[derive(PartialEq, Eq, Clone, Debug)] pub struct ListNode { pub val: i32, pub next: Option<Box<ListNode>>, } impl ListNode { #[inline] fn new(val: i32) -> Self { ListNode { next: None, val, } } } pub struct Solution {} impl Solution { /* Solution: recurtion, like merge sort; Time:O(n), Space:O(n) */ pub fn merge_two_lists(list1: Option<Box<ListNode>>, list2: Option<Box<ListNode>>) -> Option<Box<ListNode>> { match (list1, list2) { (None, None) => None, (a, None) => a, (None, b) => b, //Some(listNode): get the value from list1 or list2 which is ref of ListNode (Some(mut listNode), Some(mut listNode2)) => { if (listNode.val < listNode2.val) { listNode.next = Self::merge_two_lists(listNode.next,Some(listNode2)); return Some(listNode); } else { listNode2.next = Self::merge_two_lists(Some(listNode),listNode2.next); return Some(listNode2); } } } } }