【37】21. Merge Two Sorted Lists

21. Merge Two Sorted Lists

• Total Accepted: 190954
• Total Submissions: 500570
• Difficulty: Easy
• Contributors: Admin

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.

Solution 1: 

新建一个链表，把两个链表中较小的一个链到新链表上。

/**
 * 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* dummy = new ListNode(-1);
        ListNode* curr = dummy;
        while(l1 && l2){
            if(l1 -> val < l2 -> val){
                curr -> next = l1;
                l1 = l1 -> next;
                curr = curr -> next;
            }else{
                curr -> next = l2;
                l2 = l2 -> next;
                curr = curr -> next;
            }
        }
        if(l1){
            curr -> next = l1;
        }
        if(l2){
            curr -> next = l2;
        }
        return dummy -> next;
    }
};

Solution 2: recursive

除了base case：l1 或l2有一个为空的情况下，就返回另一个。

核心是比较l1与l2的值，如果l1的值小于l2，则对于l1 -> next 和l2调用递归函数，把返回值赋予l1 -> next; 反之，同理。

/**
 * 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) {
        if(!l1) return l2;//base case
        if(!l2) return l1;//base case
        if(l1 -> val < l2 -> val){
            //return mergeTwoLists(l1 -> next, l2);
            l1 -> next = mergeTwoLists(l1 -> next, l2);
            return l1;
        }else{
            l2 -> next = mergeTwoLists(l1, l2 -> next);
            return l2;
        }
    }
};