LeetCode:21MergeTwoSortedList

题目描述：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.

出错：访问空指针    member access within null pointer of type 'struct ListNode'

解决：优先考虑指针为空的情况避免访问出错

错误代码：

/**
 * 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 rlt(-1);
        ListNode *crt=&rlt;
        while(l1!=NULL||l2!=NULL){
            if(l1==NULL||l1->val>l2->val){
                crt->next=l2;
                l2=l2->next;
            }
            else if(l2==NULL||l1->val<l2->val){
                crt->next=l1;
                l1=l1->next;
            }
                crt=crt->next;
        }
        return rlt.next;
    }
};

其中第15行可能出现访问l2时l2文空指针的情况

解决方案如下：

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        ListNode rlt(-1);
        ListNode *crt=&rlt;
        while(l1!=NULL||l2!=NULL){
            if(l1==NULL){
                crt->next=l2;
                l2=l2->next;
            }
            else if(l2==NULL){
                crt->next=l1;
                l1=l1->next;
            }
            else if(l1->val>l2->val){
                crt->next=l2;
                l2=l2->next;
            }
            else{
                crt->next=l1;
                l1=l1->next;
            }
                crt=crt->next;
        }
        return rlt.next;
    }
};

或者：

class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode dummy(INT_MIN);
        ListNode *tail = &dummy;
        
        while (l1 && l2) {
            if (l1->val < l2->val) {
                tail->next = l1;
                l1 = l1->next;
            } else {
                tail->next = l2;
                l2 = l2->next;
            }
            tail = tail->next;
        }

        tail->next = l1 ? l1 : l2;
        return dummy.next;
    }
};