92. Reverse Linked List II

题目：

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.

链接：  http://leetcode.com/problems/reverse-linked-list-ii/

6/4/2017

注意

1. m和n的循环判断条件不一样，m是先－再判断，n是先判断再－

2. 很多断点处需要额外用ListNode varaible来标记，比如，prev, second, dummy2

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prev = dummy;

        n -= m;
        while (--m > 0) {
            prev = prev.next;
        }
        ListNode cur = prev.next;
        prev.next = null;
        ListNode dummy2 = new ListNode(-1);
        ListNode second = cur;

        while (n >= 0) {
            ListNode next = cur.next;
            cur.next = dummy2.next;
            dummy2.next = cur;
            cur = next;
            n--;
        }
        second.next = cur;
        prev.next = dummy2.next;
        return dummy.next;
    }
}

别人的答案

非常清晰的写法

学习的地方：

1. 第6行用for循环不是while循环

2. 第二个循环里，start是不变的，start.next一直在变，then总是指向当前结束时start.next，prev.next是当前的then

https://discuss.leetcode.com/topic/8976/simple-java-solution-with-clear-explanation

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;
    
}

更多讨论

https://discuss.leetcode.com/category/100/reverse-linked-list-ii