题目描述
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。
【思路1】递归
/*
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;
else if(pHead2 == NULL)
return pHead1;
ListNode* res = NULL;
if(pHead1->val <= pHead2->val){
res = pHead1;
res->next = Merge(pHead1->next, pHead2);
}else{
res = pHead2;
res->next = Merge(pHead1,pHead2->next);
}
return res;
}
};
【思路2】非递归，新建一个链表并保存头结点，将原来两个链表进行比较按顺序插入到新链表中，最后将有剩余的链表直接接上。
/*
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;
else if(pHead2 == NULL)
return pHead1;
ListNode* res = NULL;
ListNode* cur = NULL;
while(pHead1 != NULL && pHead2 != NULL){
if(pHead1->val <= pHead2->val){
if(res == NULL)
res = cur = pHead1;
else{
cur->next = pHead1;
cur = cur->next;
}
pHead1 = pHead1->next;
}else{
if(res == NULL)
res = cur = pHead2;
else{
cur->next = pHead2;
cur = cur->next;
}
pHead2 = pHead2->next;
}
}
if(pHead1 == NULL){
cur->next = pHead2;
}
if(pHead2 == NULL){
cur->next = pHead1;
}
return res;
}
};