力扣算法题—092反转链表2
反转从位置 m 到 n 的链表。请使用一趟扫描完成反转。
说明:
1 ≤ m ≤ n ≤ 链表长度。
示例:
输入: 1->2->3->4->5->NULL, m = 2, n = 4 输出: 1->4->3->2->5->NULL
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 }
