https://leetcode.com/PRoblems/reverse-linked-list-ii/?tab=Description
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.
https://discuss.leetcode.com/category/100
我的思路是用一个stack，但是效率应该比较低，所以看了一眼discuss，最高位的答案用了四个指针dummy，pre，start，then，我照着他的思路重写了一遍。
Simply just reverse the list along the way using 4 pointers: dummy, pre, start, then
public ListNode reverseBetween(ListNode head, int m, int n) { if(head == null) return null; ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list dummy.next = head; ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing for(int i = 0; i<m-1; i++) pre = pre.next; ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed ListNode then = start.next; // a pointer to a node that will be reversed // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3 // dummy-> 1 -> 2 -> 3 -> 4 -> 5 for(int i=0; i<n-m; i++) { start.next = then.next; then.next = pre.next; pre.next = then; then = start.next; } // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4 // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish) return dummy.next;}新闻热点
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