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带你快速刷完67道剑指offer

No15、反转链表

牛客网原题链接

题目描述

输入一个链表，反转链表后，输出新链表的表头。

示例1

输入

{1,2,3}

返回值

{3,2,1}

很好的解答

https://blog.csdn.net/qq_42351880/article/details/88637387

1、头插法 很经典的做法啊

struct ListNode {
    int val;
    struct ListNode* next;
    ListNode(int x) :
        val(x), next(NULL) {
    }
}; 

ListNode* ReverseList(ListNode* pHead) {

    struct ListNode* Head = NULL;
    struct ListNode* node = (ListNode*)malloc(sizeof(struct ListNode));

    while (pHead != nullptr) {
        node = pHead;
        pHead = pHead->next;

        node->next = Head;
        Head = node;
    }
    return Head;
}

void test02()
{
    ListNode* head = (ListNode*)malloc(sizeof(ListNode));
    head->val = 1;

    ListNode* node1 = (ListNode*)malloc(sizeof(ListNode));
    node1->val = 2;

    ListNode* node2 = (ListNode*)malloc(sizeof(ListNode));
    node2->val = 3;

    ListNode* node3 = (ListNode*)malloc(sizeof(ListNode));
    node3->val = 4;

    head->next = node1;
    node1->next = node2;
    node2->next = node3;
    node3->next = nullptr;

    auto node = ReverseList(head);
    while(node!=nullptr){
    
        cout << node->val << endl;
        node = node->next;
    }
    }

2、第二种方法，借助三个结点进行不断更替

ListNode* ReverseList(ListNode* pHead) {

    struct ListNode* node0 = NULL;
    struct ListNode* node1 = (ListNode*)malloc(sizeof(struct ListNode));
    struct ListNode* node2 = (ListNode*)malloc(sizeof(struct ListNode));
    node1 = pHead;
    node2 = pHead->next;
    while (node1 != nullptr) {
        node1->next = node0;

        node0 = node1;
        node1 = node2;
        if (node2!= nullptr) {
            node2 = node2 -> next;
        }
    }
    return node0;
}

二刷：

1、头插法来做，将元素开辟在栈上，这样会避免内存泄露

运行时间：3ms 占用内存：364k

ListNode* ReverseList(ListNode* pHead) {

    // 头插法
    if (pHead == nullptr || pHead->next == nullptr) return pHead;
    ListNode dummyNode = ListNode(0);
    ListNode* pre = &(dummyNode);
    pre->next = pHead;
    ListNode* cur = pHead->next;
    pHead->next = nullptr;
    //pre = cur;
    ListNode* temp = nullptr;
    while (cur != nullptr) {
        temp = cur;
        cur = cur->next;
        temp->next = pre->next;
        pre->next = temp;
    }
    return dummyNode.next;

}

2、借助三个节点来进行迭代即可

运行时间：3ms 占用内存：504k

    ListNode* ReverseList(ListNode* pHead) {


        if (pHead == nullptr || pHead->next == nullptr) return pHead;
        ListNode * pre = nullptr,*cur = pHead,*after = pHead->next;
        while(cur != nullptr){//建议画个图看看就知道了
            cur->next = pre;
            pre = cur;
            cur = after;
            if(after != nullptr)
                after = after->next;
        }
        
        return pre;
    }