92. Reverse Linked List II

92. Reverse Linked List II

 

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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.

分析： 这道题我用的方法比较麻烦，先是找出n和m的位置，然后将节点位置保存，再去将其转化为部分单链表反转的问题，最后把之前保存的位置信息再加上。

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;    }