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The basic idea is as follows:
new_head that points to head and use it to locate the immediate node before them-th (notice that it is 1-indexed) node pre;cur to be the immediate node after pre and at each time move the immediate node after cur (named move) to be the immediate node after pre. Repeat it for n - m times.1 class Solution { 2 public: 3 ListNode* reverseBetween(ListNode* head, int m, int n) { 4 ListNode* new_head = new ListNode(0); 5 new_head -> next = head; 6 ListNode* pre = new_head; 7 for (int i = 0; i < m - 1; i++) 8 pre = pre -> next; 9 ListNode* cur = pre -> next; 10 for (int i = 0; i < n - m; i++) { 11 ListNode* move = cur -> next; 12 cur -> next = move -> next; 13 move -> next = pre -> next; 14 pre -> next = move; 15 } 16 return new_head -> next; 17 } 18 };
[LeetCode] Reverse Linked List II
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4777400.html
