链表原地翻转

  1. 带有头结点的单链表L
  2. 单链表就地逆置，要求辅助空间复杂度为O(1)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
#include <iostream>
using namespace std;

typedef struct Node {
    int data;
    struct Node *next;
} Node, *LinkList;

//头插法建立链表（有头结点）
void createLinkList(LinkList &L, int n) {
    L = new Node;
    L->next = NULL;
    for (int i = 0; i < n; ++i) {
        Node *p = (Node *)malloc(sizeof(Node));
        cin >> p->data;
        p->next = NULL;
        
        p->next = L->next;
        L->next = p;
    }
}

//链表数据打印
void print(LinkList L) {
    Node *p = L->next;
    while (p != NULL) {
        //cout << p->data << ":" << p->next << " ";
        cout << p->data << " ";
        p = p->next;
    }
    cout << endl;
}

//利用头插法实现链表逆置
void reverse(LinkList &L) {
    if (L == NULL) return;
    Node *p = L->next;
    L->next = NULL;
    Node *r;//用于防止掉链
    while (p != NULL) {
        r = p->next;
        p->next = L->next;
        L->next = p;
        p = r;
    }
}

int main() {
    int n;
    cout << "enter the length of this linklist : "; cin >> n;
    LinkList L; createLinkList(L, n);
    cout << "current linklist: "; print(L);
    reverse(L);
    cout << "reverse LinkList : "; print(L);
    return 0;
}