练习:合并两个排序链表
一、题目要求:
输入两个单调递增的链表,输出两个链表合成后的链表,当然我们需要合成后的链表满足单调不减规则。
二、代码(2018–8-10)
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1 == nullptr) return pHead2;
if(pHead2 == nullptr) return pHead1;
ListNode* p1 = pHead1, *p2 = pHead2;
ListNode* h = new ListNode(-1);
ListNode* p = h;
while(p1 && p2) {
if(p1->val < p2->val) {
p->next = p1;
p = p->next;
p1 = p1->next;
}
else {
p->next = p2;
p = p->next;
p2 = p2->next;
}
}
p->next = p1 ? p1 : p2;
ListNode* head = h->next;
delete h;
return head;
}
};
代码(2018-01-16)
/*
struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class Solution {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1 == NULL) return pHead2;
if(pHead2 == NULL) return pHead1;
ListNode *head = NULL;
if(pHead1->val <= pHead2->val){
head = pHead1;
pHead1 = pHead1->next;
}
else{
head = pHead2;
pHead2 = pHead2->next;
}
ListNode *p = head;
while(pHead1 && pHead2){
if(pHead1->val <= pHead2->val){
p->next = pHead1;
pHead1 = pHead1->next;
}
else{
p->next = pHead2;
pHead2 = pHead2->next;
}
p = p->next;
}
if(pHead1) p->next = pHead1;
else if(pHead2) p->next = pHead2;
return head;
}
};
说明
在创建链表、删除链表、增加节点、删除节点时候,创建一个头节点可以方便操作。