反转单链表和双链表

Problem:
　　反转单向和双向链表
　　【题目】 分别实现反转单向链表和反转双向链表的函数。
　　【要求】 如果链表长度为N，时间复杂度要求为O(N)，额外空间
　　复杂度要求为O(1)

Solution:
　　学会使用指针

Code:

  1 #pragma once
  2
  3 #include <iostream>
  4
  5 using namespace std;
  6
  7 struct Node
  8 {
  9     int val;
 10     Node* next;
 11     Node(int a = 0) :val(a), next(NULL) {}
 12 };
 13
 14 void ReverSingleList(Node*& head)//反转单向链表
 15 {
 16     if (head->next == NULL || head->next->next == NULL)
 17         return;
 18     Node *p1, *p2, *p3;
 19     p1 = head->next;
 20     p2 = p1->next;
 21     p3 = p2->next;
 22     p1->next = NULL;
 23     while (p3)
 24     {
 25         p2->next = p1;
 26         p1 = p2;
 27         p2 = p3;
 28         p3 = p3->next;
 29     }
 30     p2->next = p1;
 31     head->next = p2;
 32 }
 33
 34
 35 struct DNode
 36 {
 37     int val;
 38     DNode* next;
 39     DNode* pre;
 40     DNode(int a = 0) :val(a), next(NULL),pre(NULL) {}
 41 };
 42
 43 void ReverDoubleList(DNode*& head)//反转双向链表，方法与单向链表一样
 44 {
 45     if (head->next == NULL || head->next->next == NULL)
 46         return;
 47     DNode *p1, *p2, *p3;
 48     p1 = head->next;
 49     p2 = p1->next;
 50     p3 = p2->next;
 51     p1->next = NULL;
 52     p1->pre = p2;
 53     while (p3)
 54     {
 55         p2->next = p1;
 56         p1->pre = p2;
 57         p2->pre = p3;
 58         p1 = p2;
 59         p2 = p3;
 60         p3 = p3->next;
 61     }
 62     p2->next = p1;
 63     p2->pre = head;
 64     head->next = p2;
 65 }
 66
 67 void Test()
 68 {
 69     int a[6] = { 1,2,3,4,5,6 };
 70     Node* head = new Node(-1);
 71     Node* p = head;
 72     for (auto n : a)
 73     {
 74         Node* q = new Node(n);
 75         p->next = q;
 76         p = q;
 77     }
 78     p->next = NULL;
 79
 80     p = head->next;
 81     cout << "原链表为： ";
 82     while (p)
 83     {
 84         cout << p->val << "->";
 85         p = p->next;
 86     }
 87
 88     ReverSingleList(head);
 89     p = head->next;
 90     cout << endl << "*******************" << endl << "反转后的链表为： ";
 91     while (p)
 92     {
 93         cout << p->val << "->";
 94         p = p->next;
 95     }
 96     cout << endl << "=============================" << endl;
 97     cout << "=============================" << endl;
 98     cout << "=============================" << endl;
 99
100
101     int b[6] = { 1,2,3,4,5,6 };
102     DNode* head2 = new DNode(-1);
103     DNode* p2 = head2;
104     for (auto n : b)
105     {
106         DNode* q2 = new DNode(n);
107         p2->next = q2;
108         q2->pre = p2;
109         p2 = q2;
110     }
111     p2->next = NULL;
112
113     p2 = head2->next;
114     cout << "原链表为的顺为： ";
115     while (p2->next)
116     {
117         cout << p2->val << "->";
118         p2 = p2->next;
119     }
120     cout << p2->val << "->";
121     cout << endl << "*******************" << endl << "原链表为的逆为： ";
122     while (p2->pre)
123     {
124         cout << p2->val << "->";
125         p2 = p2->pre;
126     }
127
128     ReverDoubleList(head2);
129     p2 = head2->next;
130     cout << endl << "*******************" << endl << "反转后的双向链表的顺为： ";
131     while (p2->next)
132     {
133         cout << p2->val << "->";
134         p2 = p2->next;
135     }
136     cout << p2->val << "->";
137     cout << endl << "*******************" << endl << "反转后的双向链表的逆为： ";
138     while (p2->pre)
139     {
140         cout << p2->val << "->";
141         p2 = p2->pre;
142     }
143     cout << endl << "=============================" << endl;
144 }

原文地址：https://www.cnblogs.com/zzw1024/p/10989154.html