题解 | #合并两个排序的链表#
合并两个排序的链表
https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
ListNode* r = (ListNode*)malloc(sizeof(ListNode));
r->next = NULL;
ListNode* newList = (ListNode*)malloc(sizeof(ListNode));
newList = r;
while(pHead1 != NULL && pHead2 != NULL){
ListNode* p = (ListNode*)malloc(sizeof(ListNode));
if(pHead1->val > pHead2->val){
p->val = pHead2->val;
pHead2 = pHead2->next;
}else{
p->val = pHead1->val;
pHead1 = pHead1->next;
}
r->next = p;
r = r->next;
}
while(pHead1){
ListNode* p = (ListNode*)malloc(sizeof(ListNode));
p->val = pHead1->val;
r->next = p;
r = r->next;
pHead1 = pHead1->next;
}
while(pHead2){
ListNode* p = (ListNode*)malloc(sizeof(ListNode));
p->val = pHead2->val;
r->next = p;
r = r->next;
pHead2 = pHead2->next;
}
return newList->next;
}
};
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
ListNode* r = (ListNode*)malloc(sizeof(ListNode));
r->next = NULL;
ListNode* newList = (ListNode*)malloc(sizeof(ListNode));
newList = r;
while(pHead1 != NULL && pHead2 != NULL){
ListNode* p = (ListNode*)malloc(sizeof(ListNode));
if(pHead1->val > pHead2->val){
p->val = pHead2->val;
pHead2 = pHead2->next;
}else{
p->val = pHead1->val;
pHead1 = pHead1->next;
}
r->next = p;
r = r->next;
}
while(pHead1){
ListNode* p = (ListNode*)malloc(sizeof(ListNode));
p->val = pHead1->val;
r->next = p;
r = r->next;
pHead1 = pHead1->next;
}
while(pHead2){
ListNode* p = (ListNode*)malloc(sizeof(ListNode));
p->val = pHead2->val;
r->next = p;
r = r->next;
pHead2 = pHead2->next;
}
return newList->next;
}
};