21. Merge Two Sorted Lists 混合插入有序链表

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

Difficulty: Easy

这道题让我们把两个有序的链表合并成一个有序的链表,与剑指offer 16题完全相同.

解法1

按照剑指offer 16题的递归解法

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

解法2

采用循环的方式

/**
 * 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), *cur = dummy;
        while(l1&&l2)
        {
            if (l1->val < l2->val){
                cur->next = l1;
                l1 = l1->next;
            }
            else {
                cur->next = l2;
                l2 = l2->next;
            }
            cur = cur->next;
        }
        cur->next = l1?l1:l2;
        //要记得释放没用的节点
        ListNode* res = dummy->next;
        delete dummy;
        dummy = NULL;
        return res;
    }
};
 
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