力扣算法题—092反转链表2

 1 #include "_000库函数.h"
 2 
 3 
 4 struct ListNode {
 5     int val;
 6     ListNode *next;
 7     ListNode(int x) : val(x), next(NULL) {}
 8 };
 9 
10 
11 //只能扫描一遍
12 //将中间要反转的数取出来在放入链表中
13 class Solution {
14 public:
15     ListNode* reverseBetween(ListNode* head, int m, int n) {
16         if (!head || m <= 0 || n <= 0 || n <= m)return head;
17         //加头
18         ListNode*p = new ListNode(-1);
19         p->next = head;
20         head = p;
21 
22         stack<int>s;
23         ListNode*pre = new ListNode(0);
24         while (p && n >0) {
25             --m;
26             if (m == 0)pre = p;            
27             p = p->next;
28             if (p && m <= 0)
29                 s.push(p->val);
30             --n;
31         }
32         if (p == NULL)return head->next;//m,n超过了链表长度
33         pre->next = p->next;
34         while (!s.empty()) {
35             ListNode*q = new ListNode(0);
36             q->val = s.top();
37             q->next = pre->next;            
38             pre->next = q;
39             pre = q;
40             s.pop();
41         }
42         return head->next;
43     }
44 };
45 
46 //走一步反转一个数据
47 //不用对m,n的大小进行判断
48 class Solution {
49 public:
50     ListNode *reverseBetween(ListNode *head, int m, int n) {
51         ListNode *dummy = new ListNode(-1), *pre = dummy;
52         dummy->next = head;
53         for (int i = 0; i < m - 1; ++i) pre = pre->next;
54         ListNode *cur = pre->next;
55         for (int i = m; i < n; ++i) {//用交换法
56             ListNode *t = cur->next;
57             cur->next = t->next;
58             t->next = pre->next;
59             pre->next = t;
60         }
61         return dummy->next;
62     }
63 };
64 
65 
66 void T092() {
67     Solution s;
68     vector<int>v;
69     ListNode *head = new ListNode(0);
70     ListNode *p = head;
71     v = { 1,2,3,4,5 };
72     for (auto a : v) {
73         ListNode *q = new ListNode(0);
74         q->val = a;
75         p->next = q;
76         p = q;
77     }
78     p = head->next;
79     while (p) {
80         cout << p->val << "->";
81         p = p->next;
82     }
83     cout << endl;
84 
85 
86     p = s.reverseBetween(head->next, 2, 6);
87     while (p) {
88         cout << p->val << "->";
89         p = p->next;
90     }
91     cout << endl;
92 }