大厂笔试题
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#include<iostream>
using namespace std;
struct ListNode
{
int val;
ListNode* next;
};
struct ListNode* Merge(struct ListNode* pHead1, struct ListNode* pHead2 ) {
// write code here
if(pHead1 == NULL)return pHead2;
if(pHead2 == NULL)return pHead1;
struct ListNode* p1 = (pHead1->val <= pHead2->val ? pHead1 : pHead2);
struct ListNode* p2 = (pHead1->val > pHead2->val ? pHead1 : pHead2);
struct ListNode* p = p1; //p1是主链
// temp存放中间指针,有可能在p1中有可能在p2中,最终每次指向p1链两个node之间和大node最近的那个点
struct ListNode* temp =NULL;
while((p1 != NULL)&&(p2 != NULL)){
if(p1->val <= p2->val){
temp = p1;
p1 = p1->next;
}
else {
temp->next = p2;
temp = p2;
p2 = p2->next;
temp->next = p1;
}
}
if(p1 == NULL){
temp->next =p2;
return p;
}else return p;
}
int main(){
ListNode* one = new ListNode;
one->val = 1;
ListNode* two = new ListNode;
two->val = 3;
ListNode* three = new ListNode;
three->val = 4;
ListNode* one2 = new ListNode;
one2->val = 1;
ListNode* two2 = new ListNode;
two2->val = 2;
ListNode* three2 = new ListNode;
three2->val = 5;
one->next = two;
two->next = three;
three->next = nullptr;
one2->next = two2;
two2->next = three2;
three2->next = nullptr;
ListNode* res = Merge(one, one2);
ListNode* cur = res;
cout << "init success" << endl;
while(cur != nullptr){
cout << cur->val << endl;
cur = cur->next;
}
return 0;
}