21.Merge-Two-Sorted-Lists

21. Merge Two Sorted Lists

题目地址

题目描述

Merge two sorted (ascending) linked lists and return it as a new sorted list. The new sorted list should be made by splicing together the nodes of the two lists and sorted in ascending order.

代码

Approach 1: Iteration

Intuition
We can achieve the same idea via iteration by assuming that l1 is entirely less than l2 and processing the elements one-by-one, inserting elements of l2 in the necessary places in l1.
Complexity Analysis
  • Time complexity : O(n + m)
  • Space complexity : O(1)
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode curr = dummy;
while (l1 != null && l2 != null) {
if (l1.val > l2.val){
curr.next = l2;
l2 = l2.next;
} else {
curr.next = l1;
l1 = l1.next;
}
curr = curr.next;
}
curr.next = (l1 != null) ? l1 : l2;
return dummy.next;
}
}

Approach 2: Recursion

Intuition
We can recursively define the result of a merge operation on two lists as the following (avoiding the corner case logic surrounding empty lists):
list1.next = merge(list1.next, list2) list1.val < list2.val
list2.next = merge(list1, list2.next) otherwise
Namely, the smaller of the two lists' heads plus the result of a merge on the rest of the elements.
Complexity Analysis
  • Time complexity : O(n + m)
  • Space complexity : O(n + m)
    The first call to mergeTwoLists does not return until the ends of both l1 and l2 have been reached, so n + m stack frames consume O(n + m)space.
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
} else if (l2 == null) {
return l1;
} else if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
}