[LeetCode] 92. Reverse Linked List II

Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.
Example1:

Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]

Example2:

Input: head = [5], left = 1, right = 1
Output: [5]

这道题思考很简单，就是找到最开始需要反转的前一个点和第一个点记下来， 然后反转，再把反转的最后一个点以及反转后面的第一个点记下来，最后串联起来。需要注意边界，比如第一个点就是需要反转的点，所以设置一个dummy node会让整个过程简化。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseBetween(ListNode head, int left, int right) {
        if (head == null || left >= right || head.next == null) {
            return head;
        }
       //寻找需要反转的第一个点以及之前的点。之前的点用ps记住，第一个点用start记住。注意start不要指向head，要指向ps.next
        ListNode dummy = new ListNode(0, head);
        ListNode ps = dummy;
        int step  = right - left + 1;
        
        while (--left > 0) {
            ps = ps.next;
            head = head.next;
        }
        ListNode start = ps.next;
        ListNode prev = ps;
        ListNode next = null;
        
       //反转需要反转的list，之后prev记住的是反转的最后一个点，head是反转后的第一个点
        while (step-- > 0 && head != null) {
            next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }

      //把该串联的三块串联号
        ps.next = prev;
        start.next = head;
        return dummy.next;
    }     
}