Merge Two Sorted Lists

描述

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.

分析

无

代码

Python

// Merge Two Sorted Lists
// 时间复杂度O(min(m,n))，空间复杂度O(1)
public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        for (ListNode p = dummy; l1 != null || l2 != null; p = p.next) {
            int val1 = l1 == null ? Integer.MAX_VALUE : l1.val;
            int val2 = l2 == null ? Integer.MAX_VALUE : l2.val;
            if (val1 <= val2) {
                p.next = l1;
                l1 = l1.next;
            } else {
                p.next = l2;
                l2 = l2.next;
            }
        }
        return dummy.next;
    }
}

Java

// Merge Two Sorted Lists
// 时间复杂度O(min(m,n))，空间复杂度O(1)
class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode dummy(-1);
        for (ListNode* p = &dummy; l1 != nullptr || l2 != nullptr; p = p->next) {
            int val1 = l1 == nullptr ? INT_MAX : l1->val;
            int val2 = l2 == nullptr ? INT_MAX : l2->val;
            if (val1 <= val2) {
                p->next = l1;
                l1 = l1->next;
            } else {
                p->next = l2;
                l2 = l2->next;
            }
        }
        return dummy.next;
    }
};

C++

# Merge Two Sorted Lists
# Time complexity O(min(m,n)), Space complexity O(1)
class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode(-1)
        p = dummy
        while l1 or l2:
            val1 = float('inf') if not l1 else l1.val
            val2 = float('inf') if not l2 else l2.val
            if val1 <= val2:
                p.next = l1
                l1 = l1.next
            else:
                p.next = l2
                l2 = l2.next
            p = p.next
        return dummy.next

相关题目

• Merge Two Sorted Arrays
• Merge k Sorted Lists