【Leetcode 21】合并两个有序链表

题目描述

java版本

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;
        }

    }
}

python版本

class Solution:
    def mergeTwoLists(self, l1, l2):
        if l1 is None:
            return l2
        elif l2 is None:
            return l1
        elif l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)   # l1的值比l2小，所以l1向后移动

            return l1        #返回较小的那个节点，此时，l1小就返回l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2

方法2 ：4个指针

注意哨兵节点的使用，没有哨兵节点，代码就会复杂一点

java版本

class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode prehead = new ListNode(-1);

        ListNode prev = prehead;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                prev.next = l1;
                l1 = l1.next;
            } else {
                prev.next = l2;
                l2 = l2.next;
            }
            prev = prev.next;
        }

        // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
        prev.next = l1 == null ? l2 : l1;

        return prehead.next;
    }
}

python版本

class Solution:
    def mergeTwoLists(self, l1, l2):
        # maintain an unchanging reference to node ahead of the return node.
        prehead = ListNode(-1)

        prev = prehead
        while l1 and l2:
            if l1.val <= l2.val:
                prev.next = l1
                l1 = l1.next
            else:
                prev.next = l2
                l2 = l2.next            
            prev = prev.next

        # exactly one of l1 and l2 can be non-null at this point, so connect
        # the non-null list to the end of the merged list.
        prev.next = l1 if l1 is not None else l2

        return prehead.next

参考链接

与剑指offer题目相同​

​​https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/he-bing-liang-ge-you-xu-lian-biao-by-leetcode-solu/​​