反转链表

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

struct LinkedListNode{
    int value;
    LinkedListNode *pNext;
    LinkedListNode(int v) : value(v) {}
};

//创建链表
LinkedListNode *CreateLinkedList(int arr[], int len)
{
    LinkedListNode *pHead = new LinkedListNode(arr[0]);
    LinkedListNode *pPre = pHead;
    LinkedListNode *pCurr = NULL;

    //每次进入该循环的时候，pPre是起作用的
    //pCurr都会重新构造一个LinkedListNode
    for (int i = 1; i < len; i++)
    {
        pCurr = new LinkedListNode(arr[i]);
        pCurr->pNext = NULL;
        pPre->pNext = pCurr;
        pPre = pCurr;
    }
    return pHead;
}

//输出链表
void PrintLinkedList(LinkedListNode *pHead)
{
    while (pHead)
    {
        cout<<pHead->value<<endl;
        pHead = pHead->pNext;
    }
}

//反转链表
//需要3个指针，pCurr指向正在改变的Node，pPre指向需要改变到的位置，pNex记录不再被pCurr指向的下一个位置
LinkedListNode *ReverseLinkedList(LinkedListNode *pHead)
{
    if (!pHead)
        return NULL;

    LinkedListNode *pPre = NULL;
    LinkedListNode *pCurr = pHead;
    LinkedListNode *pNex = pHead->pNext;
    while (pCurr)
    {
        pCurr->pNext = pPre;
        pPre = pCurr;
        pCurr = pNex;
        if (pNex)//当pCurr为最后一个元素的时候，pNex已经为NULL，所以需要check
            pNex = pNex->pNext ? pNex->pNext : NULL;
    }
    return pPre;
}

int main()
{
    int arr[10] = {4, 2, 5, 7, 4, 1, 7, 3, 1, 9};
    LinkedListNode *pHead = CreateLinkedList(arr, 10);
    PrintLinkedList(pHead);
    pHead = ReverseLinkedList(pHead);
    cout<<endl<<"After Reverse:"<<endl;
    PrintLinkedList(pHead);
    return 0;
}

EOF