Linked List-Easy怎么将两个排序的链表合并

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将两个排序的链表合并，返回一个新链表，返回的新链表也是排好序的。

解题思路：

• 创建两个链表，一个负责保存头节点，一个负责记录比较后的结果。

Language : c

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {struct ListNode* newlist = (struct ListNode *)malloc(sizeof(struct ListNode));struct ListNode* temp = (struct ListNode *)malloc(sizeof(struct ListNode));    newlist = temp;while(l1 && l2){if(l1->val < l2->val){            temp->next = l1;            l1 = l1->next;        }else{            temp->next = l2;            l2 = l2->next;        }        temp = temp->next;    }    temp->next = l1 ? l1 : l2;return newlist->next;}

Language : cpp

/** * 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 newlist(INT_MIN);        ListNode *temp = &newlist;if(l1 == NULL && l2 == NULL){            return NULL;        }if(l1 != NULL && l2 == NULL){            return l1;        }if(l1 == NULL && l2 != NULL){            return l2;        }while(l1 && l2){if(l1->val < l2->val){                temp->next = l1;                l1 = l1->next;            }else{                temp->next = l2;                l2 = l2->next;            }            temp = temp->next;        }        temp->next = l1 ? l1 : l2;        return newlist.next;    }};

Language：python

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):def mergeTwoLists(self, l1, l2):"""        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """result = cur = ListNode(0)while l1 and l2:if l1.val < l2.val:                cur.next = l1                l1 = l1.nextelse:                cur.next = l2                l2 = l2.next            cur = cur.next        cur.next = l1 or l2return result.next

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