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# No16、合并两个有序链表
# 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。
# 示例1
输入
{1,3,5},{2,4,6}
1
返回值
{1,2,3,4,5,6}
1
# 1、常规做法,非递归花了好久才做出来
struct ListNode {
int val;
struct ListNode* next;
ListNode(int x) :
val(x), next(NULL) {
}
};
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if (pHead1 == nullptr) return pHead2;
if (pHead2 == nullptr) return pHead1;
ListNode* Head = (ListNode*)malloc(sizeof(struct ListNode));
if (pHead1->val <= pHead2->val) {
Head = pHead1;
pHead1 = pHead1->next;
}else {
Head = pHead2;
pHead2 = pHead2->next;
}
ListNode* node = (ListNode*)malloc(sizeof(struct ListNode));
node = Head;
while (pHead1 && pHead2) {
if (pHead1->val <= pHead2->val) {
node->next = pHead1;
pHead1 = pHead1->next;
node = node->next;
}
else {
node->next = pHead2;
pHead2 = pHead2->next;
node = node->next;
}
}
if (pHead1 != nullptr) {
node->next = pHead1;
}
else {
node->next = pHead2;
}
return Head;
}
void test02()
{
ListNode* head = (ListNode*)malloc(sizeof(ListNode));
head->val = 1;
ListNode* node1 = (ListNode*)malloc(sizeof(ListNode));
node1->val = 5;
ListNode* node2 = (ListNode*)malloc(sizeof(ListNode));
node2->val = 9;
ListNode* node3 = (ListNode*)malloc(sizeof(ListNode));
node3->val = 11;
//node3->next = NULL;
head->next = node1;
node1->next = node2;
node2->next = node3;
node3->next = nullptr;
ListNode* head2 = (ListNode*)malloc(sizeof(ListNode));
head2->val = 3;
ListNode* node12 = (ListNode*)malloc(sizeof(ListNode));
node12->val = 3;
ListNode* node22 = (ListNode*)malloc(sizeof(ListNode));
node22->val = 4;
ListNode* node32 = (ListNode*)malloc(sizeof(ListNode));
node32->val = 9;
//node3->next = NULL;
head2->next = node12;
node12->next = node22;
node22->next = node32;
node32->next = nullptr;
auto node = Merge(head,head2);
while(node!=nullptr){
cout << node->val << endl;
node = node->next;
}
}
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# 2、递归版本
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if (pHead1 == nullptr) return pHead2;
if (pHead2 == nullptr) return pHead1;
if (pHead1->val <= pHead2->val) {
pHead1->next = Merge(pHead1->next, pHead2);
return pHead1;
}
else {
pHead2->next = Merge(pHead1, pHead2->next);
return pHead2;
}
}
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# 二刷:很容易了
# 1、迭代版本,依然还是迭代版本要快一点
运行时间:2ms 占用内存:480k
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1 == nullptr) return pHead2;
if(pHead2 == nullptr) return pHead1;
ListNode *newHead = new ListNode(-1),*node = newHead;
//newHead->next=node;
while(pHead1 != nullptr && pHead2 != nullptr){
if(pHead1->val > pHead2->val) swap(pHead1,pHead2);
node->next = pHead1;
pHead1 = pHead1->next;
node = node->next;
}
node->next = (pHead1 ? pHead1:pHead2);
return newHead->next;
}
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# 2、递归版本
运行时间:3ms 占用内存:504k
void MergeCore(ListNode*newHead, ListNode*node1, ListNode*node2){
if(node1 == nullptr || node2 == nullptr) {
newHead->next = (node1?node1:node2);
return ;
}
if(node1->val < node2->val){
newHead->next = node1;
node1 = node1->next;
}
else{
newHead->next = node2;
node2 = node2->next;
}
newHead = newHead->next;
MergeCore(newHead,node1,node2);
}
ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
{
if(pHead1 == nullptr) return pHead2;
if(pHead2 == nullptr) return pHead1;
ListNode *newHead = new ListNode(-1),*node = newHead;
MergeCore(node, pHead1, pHead2);
return newHead->next;
}
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← No15、反转链表 No17、树的子结构 →