[LeetCode] 21. 合并两个有序链表

题目链接:https://leetcode-cn.com/problems/merge-two-sorted-lists/

题目描述:

将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例:

输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4

思路:

思路1:

迭代方法.

每次选两个链表头结点最小的,比如:我们生活中,有两个已经按照高矮排好的队伍,我们如何把变成一个队伍!当然,每次选两个队伍排头的,比较他们的高矮!组成新的的队伍.

时间复杂度:\(O(m+n)\)

空间复杂度:\(O(m+n)\)

思路2:

递归方法

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代码:

python

迭代

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        dummy = ListNode(0)
        p = dummy
        while l1 or l2:
            if l1 and l2:
                tmp1 = l1.val
                tmp2 = l2.val
                if tmp1 < tmp2:
                    p.next = ListNode(tmp1)
                    l1 = l1.next
                else:
                    p.next = ListNode(tmp2)
                    l2 = l2.next
            elif l1:
                p.next = ListNode(l1.val)
                l1 = l1.next
            elif l2:
                p.next = ListNode(l2.val)
                l2 = l2.next
            p = p.next
        return dummy.next

java

/**
 * 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) {
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (l1 != null && l2 != null){
            if (l1.val < l2.val){
                p.next = l1;
                l1 = l1.next;
            }
            else {
                p.next = l2;
                l2 = l2.next;
            }
            p = p.next;
        }
        if (l1 == null) p.next = l2;
        if (l2 == null) p.next = l1;
        return dummy.next;
    }
}

递归

python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if not l1: return l2
        if not l2: return l1
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next,l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1,l2.next)
            return l2

java

/**
 * 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;
        if (l2 == null) return l1;
        if (l1.val < l2.val) {
            l1.next = mergeTwoLists(l1.next,l2);
            return l1;
        }
        else{
            l2.next = mergeTwoLists(l1,l2.next);
            return l2;
        }
        
    }