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[LeetCode] Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode *reverseBetween(ListNode *head, int m, int n) {12 if (head == NULL || m > n) return NULL;13 ListNode* pmprev = NULL;// 始终指向第m个节点的前缀14 ListNode* pn = NULL; //指向逆置后的第n个节点15 ListNode* pm = NULL; //指向第m个节点16 ListNode *pter = NULL; //操作节点17 ListNode *pter_next = NULL; //指向操作节点的下一个节点18 pter = head;19 20 for (int i = 1; i < m; ++i) {21 pmprev = pter;22 pter = pter->next;23 }24 25 pn = pm = pter;26 for (int i = m; i <= n; ++i) {27 pter_next = pter->next;28 pter->next = pn;29 pn = pter;30 pter = pter_next;31 }32 33 if (pmprev == NULL) { //如果头结点也在逆置范围中34 head = pn;35 } else {36 pmprev->next = pn;37 }38 39 pm->next = pter_next;40 41 return head;42 }43 };
[LeetCode] Reverse Linked List II
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