带你快速刷完67道剑指offer

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}

返回值

{1,2,3,4,5,6}

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;
    }
}

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;
    }
    }

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;
    }

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;
    }