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  • 【LeetCode】21. 合并两个有序链表

    链接:

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

    描述:

    将两个升序链表合并为一个新的 升序 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

    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {}

    思路:递归

    C++

    展开后查看
    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode() : val(0), next(nullptr) {}
     *     ListNode(int x) : val(x), next(nullptr) {}
     *     ListNode(int x, ListNode *next) : val(x), next(next) {}
     * };
     */
    class Solution {
    public:
        ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
            if(l1 == nullptr){
                return l2;
            }else if(l2 == nullptr){
                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;
            }
        }
    };
    

    Java

    展开后查看
    /**
     * 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; }
     * }
     */
    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;
            }
        }
    }
    

    思路:非递归

    C++

    展开后查看
    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode() : val(0), next(nullptr) {}
     *     ListNode(int x) : val(x), next(nullptr) {}
     *     ListNode(int x, ListNode *next) : val(x), next(next) {}
     * };
     */
    class Solution {
    public:
        ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
            ListNode* dummyHead = new ListNode(-1);
            ListNode* p = dummyHead;
            while(l1 != nullptr && l2 != nullptr){
                if(l1->val <= l2->val){
                    p->next = l1;
                    l1 = l1->next;
                }else{
                    p->next = l2;
                    l2 = l2->next;
                }
                p = p->next;
            }
            p->next = l1 != nullptr ? l1 : l2;
            p = dummyHead->next;
            delete dummyHead;
            return p;
        }
    };
    

    Java

    展开后查看
    /**
     * 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; }
     * }
     */
    class Solution {
        public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
            ListNode dummyHead = new ListNode(-1);
            ListNode p = dummyHead;
            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;
            }
            p.next = l1 != null ? l1 : l2;
            return dummyHead.next;
        }
    }
    
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  • 原文地址:https://www.cnblogs.com/crazyBlogs/p/13164038.html
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