LeetCode --- 21. Merge Two Sorted Lists

21. Merge Two Sorted Lists

Difficulty: Easy

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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

Solution

思路1

使用l3链表，进行处理。

Language: C++

/**
 * 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 == NULL) return l2;
        if (l2 == NULL) return l1;
        ListNode ln(0);
        ListNode *l3 = &ln;
        while (l1 && l2){
            if (l1->val < l2->val){
                l3->next = l1;
                l1 = l1->next;
            }
            else{
                l3->next = l2;
                l2 = l2->next;
            }
            l3 = l3->next;
        }
        l3->next = l1 ? l1 : l2;
        return ln.next;
    }
};

思路2

  采用递归法，求链表！

Language: C++

/**
 * 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 == NULL) return l2;                       //一串为空情况
        if(l2 == NULL) return l1;
        if(l1->val < l2->val){                          //采用递归
            l1->next = mergeTwoLists(l1->next, l2);
            return l1;
        }
        else{
            l2->next = mergeTwoLists(l1, l2->next);
            return l2;
        }
    }
};