leetcode21合并两个排好序的单链表

方法一&＃xff1a;

/*** 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* p1&＃61;l1;ListNode* p2&＃61;l2;if(l1&＃61;&＃61;NULL){return l2;}else if(l2&＃61;&＃61;NULL){return l1;}ListNode* cur&＃61;(l1->val<&＃61;l2->val?l1:l2);ListNode* head&＃61;cur;if(l1->val<&＃61;l2->val){p1&＃61;p1->next;}else{p2&＃61;p2->next;}while(p1&&p2){if(p1->val<&＃61;p2->val){cur->next&＃61;p1;p1&＃61;p1->next;}else{cur->next&＃61;p2;p2&＃61;p2->next;}cur&＃61;cur->next;}while (p1){cur->next &＃61; p1;p1 &＃61; p1->next; cur &＃61; cur->next;}while (p2){cur->next &＃61; p2;p2 &＃61; p2->next;cur &＃61; cur->next;}return head;}
};

方法二&＃xff1a;递归

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {if (l1 &＃61;&＃61; NULL) return l2;if (l2 &＃61;&＃61; NULL) return l1;if (l1->val <&＃61; l2->val) {l1->next &＃61; mergeTwoLists(l1->next, l2);return l1;} else {l2->next &＃61; mergeTwoLists(l1, l2->next);return l2;}
}