leetcode-21-Merge-Two-Sorted-Lists

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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

合并两个有序的链表。分别采用递归和遍历的方式。

解决方案1(Python)

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

class Solution(object):
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        if l1 == None:
            return l2
        if l2 == None:
            return l1
        if l1.val <= l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        l2.next = self.mergeTwoLists(l1, l2.next)
        return l2

解决方案2(C++)

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode* head;
        if(l1 == NULL) return l2;
        if(l2 == NULL) return l1;
        
        if(l1->val <= l2->val) {
            head = l1;
            l1 = l1->next;
        }else {
            head = l2;
            l2 = l2->next;
        }
        ListNode* head_back = head;
        
        while(l1 != NULL && l2 != NULL) {
            if(l1->val <= l2->val) {
                head->next = l1;
                l1 = l1->next;
            }else {
                head->next = l2;
                l2 = l2->next;
            }
            head->next->next = NULL;
            head = head->next;
        }
        
        if(l1) {
            head->next = l1;
        }
        if(l2) {
            head->next = l2;
        }
        return head_back;
    }   
};

解决方案3(Java)

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public 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;
        }
        l2.next = mergeTwoLists(l1, l2.next);
        return l2;
    }
}

解决方案4（Golang）

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {
    if l1 == nil {
        return l2
    }
    if l2 == nil {
        return l1
    }
    var result *ListNode
    if l1.Val >= l2.Val {
        result = l2
        result.Next = mergeTwoLists(l1, l2.Next)
    } else {
        result = l1
        result.Next = mergeTwoLists(l1.Next, l2)
    }
    return result
}

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