反轉鏈表-leetcode92

給你單鏈表的頭指針 head 和兩個整數 left 和 right ，其中 left <= right 。請你反轉從位置 left 到位置 right 的鏈表節點，返回 反轉後的鏈表 。

示例 1：

輸入：head = [1,2,3,4,5], left = 2, right = 4
輸出：[1,4,3,2,5]
示例 2：

輸入：head = [5], left = 1, right = 1
輸出：[5]

//leetcode submit region begin(Prohibit modification and deletion)

/**
 * 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 || head.next == null || left == right) {
            return head;
        }


        //虛擬節點爲了避免頭節點爲空
        ListNode dummy = new ListNode(-1);
        dummy.next = head;

        //分別記錄要反轉的前節點和要反轉的第一個節點
        ListNode currDummy = dummy;
        ListNode prevDummy = null;

        ListNode prev = null;
        ListNode curr = null;

        for (int i = 0; i < left; i++) {
            prevDummy = currDummy;
            currDummy = currDummy.next;
        }
        curr = currDummy;

        //反轉範圍內的節點
        for (int i = left; i <= right && curr !=null; i++){
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }

        //拼接節點，反轉前的結點 拼反轉後的頭結點
        prevDummy.next = prev;
        //反轉後的最後一個節點拼接right後的那個節點
        currDummy.next = curr;


         return dummy.next;

    }


}
//leetcode submit region end(Prohibit modification and deletion)