sword to offer 016

Desicription

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。

Solution

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struct ListNode {
int val;
struct ListNode *next;
ListNode(int x) :
val(x), next(NULL) {
}
};*/
class {
public:
ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
ListNode* head = nullptr;
ListNode* current = nullptr;
while(pHead1 || pHead2) {
if(head == nullptr) {
if(pHead1 == nullptr) {
head = pHead2;
pHead2 = pHead2->next;
} else if(pHead2 == nullptr) {
head = pHead1;
pHead1 = pHead1->next;
} else {
if(pHead1->val <= pHead2->val) {
head = pHead1;
pHead1 = pHead1->next;
} else {
head = pHead2;
pHead2 = pHead2->next;
}
}
current = head;
} else {
if(pHead1 == nullptr) {
current->next = pHead2;
pHead2 = pHead2->next;
} else if(pHead2 == nullptr) {
current->next = pHead1;
pHead1 = pHead1->next;
} else {
if(pHead1->val <= pHead2->val) {
current->next = pHead1;
pHead1 = pHead1->next;
} else {
current->next = pHead2;
pHead2 = pHead2->next;
}
}
current = current->next;
}
}
return head;
}
};