13.反转链表

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反转链表

参与人数：5997时间限制：1秒空间限制：32768K
本题知识点：链表
算法知识视频讲解

题目描述

输入一个链表，反转链表后，输出链表的所有元素。

这道题我做了近40分钟，对链表的插入操作不熟悉！

// 12.cpp : 定义控制台应用程序的入口点。//#include "stdafx.h"#include <stack>using namespace::std;struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}};class Solution {public:ListNode* ReverseList(ListNode* pHead) {if (!pHead) return NULL;stack<ListNode*> stack;ListNode* pNode = pHead;while (pNode) {stack.push(pNode);pNode = pNode->next;}ListNode* pNew = stack.top();stack.pop();ListNode* tmp = pNew;while (!stack.empty()) {ListNode* node = stack.top();node->next = pNew->next;pNew->next = node;pNew = pNew->next;stack.pop();}//pNew = tmp;return tmp;}};int _tmain(int argc, _TCHAR* argv[]){ListNode n1(1);ListNode n2(2);ListNode n3(3);ListNode n4(4);ListNode n5(5);n1.next = &n2;n2.next = &n3;n3.next = &n4;n4.next = &n5;Solution s;s.ReverseList(&n1);return 0;}

while (!stack.empty()) {ListNode* node = stack.top();node->next = pNew->next;pNew->next = node;pNew = pNew->next;stack.pop();}

第二次做：

/*struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}};*/class Solution {public:    ListNode* ReverseList(ListNode* pHead) {if ( pHead == NULL ) return NULL ;                vector<int> vec ;        ListNode* pNode = pHead ;        while ( pNode != NULL ) {            vec.push_back( pNode -> val ) ;            pNode = pNode -> next ;        }                ListNode* pNew = new ListNode( vec[ vec.size() - 1 ] ) ;        ListNode* pCur = pNew ;                for ( int i = vec.size() - 2; i >= 0; -- i ) {            ListNode* tmp = new ListNode( vec[i] ) ;            pCur -> next = tmp ;            pCur = pCur -> next ;        }        return pNew ;    }};

while循环中我忘了

pNode = pNode -> next ;

。。。下次注意！

第三次做：

/*struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}};*/class Solution {public:    ListNode* ReverseList(ListNode* pHead) {if ( pHead == NULL ) return NULL ;                vector<int> vec ;        ListNode* pNode = pHead ;        while ( pNode != NULL ) {            vec.push_back( pNode->val ) ;            pNode = pNode->next ;        }                std::reverse( vec.begin(), vec.end() ) ;                ListNode* pNew = new ListNode( vec[0] ) ;        ListNode* pCur = pNew ;        for ( int i = 1; i < vec.size(); ++ i ) {            ListNode* tmp = new ListNode( vec[i] ) ;            pCur->next = tmp ;            pCur = pCur->next ;        }                return pNew ;    }};

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