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。
/*
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* pMergedHead = NULL;
ListNode* pH=NULL;
if(pHead1->val < pHead2->val)//确定头节点
{
pMergedHead = pHead1;
pHead1=pHead1->next;
}
else
{
pMergedHead = pHead2;
pHead2=pHead2->next;
}
pH=pMergedHead;//备份头结点
while(pHead1&&pHead2)
{
if(pHead1->val < pHead2->val)
{
pMergedHead->next = pHead1;
pHead1=pHead1->next;
pMergedHead=pMergedHead->next;
}
else
{
pMergedHead->next=pHead2;
pHead2=pHead2->next;
pMergedHead=pMergedHead->next;
}
}
if(pHead1==nullptr||pHead2==nullptr)//有一个链表走到尾,把没走到尾的链表连接,避免断链情况
{
if(pHead1==nullptr)
pMergedHead->next=pHead2;
if(pHead2==nullptr)
pMergedHead->next=pHead1;
}
return pH;
}
};