剑指 Offer 25. 合并两个排序的链表

原题链接：剑指 Offer 25. 合并两个排序的链表

 

solution:

        归并的思想

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