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0021. Merge Two Sorted Lists
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0021. Merge Two Sorted Lists
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""" 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]
"""
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional:
# recursively Time:O(n+m) Space:O(n+m)
if list1 is None:
return list2
elif list2 is None:
return list1
elif list1.val < list2.val:
list1.next = self.mergeTwoLists(list1.next, list2)
return list1
else:
list2.next = self.mergeTwoLists(list1, list2.next)
return list2
# Iteritive Time:O(n+m) Space:O(1)
dummyHead = ListNode(0) # node in result list
cur = dummyHead # pointer in result list
while list1 and list2:
if list1.val <= list2.val:
cur.next = list1
list1 = list1.next
else:
cur.next = list2
list2 = list2.next
cur = cur.next
# l1 or l2 may still have nodes left
cur.next = list1 if list1 else list2 # connect non-null nodes left
return dummyHead.next
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
// O(n + m) O(n + m)
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;
}
}
}