# [LeetCode 92. Reverse Linked List II ](https://leetcode.com/explore/challenge/card/june-leetcoding-challenge-2021/606/week-4-june-22nd-june-28th/3789/)
### Daily challenge Jun 23, 2021 (EASY)

>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.


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

:::info
**Example 2:**
**Input:** head = [5], left = 1, right = 1
**Output:** [5]
:::

:::warning
**Constraints:**
* The number of nodes in the list is n.
* 1 <= n <= 500
* -500 <= Node.val <= 500
* 1 <= left <= right <= n
:::

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### Approach 1 : Dummy Node :book:
**`0 ms ( % )`** **`O()`**

使 **`dummy`** 指向 **`head`**，另外宣告三個 pointer **`( p, start, end )`** 分別表示 `node before start` 、 `a pointer to the beginning of a sub-list that will be reversed` 、 `a pointer to a node that will be reversed`，接下來進行 reverse。

        for(int i=left; i<right; i++){
            start->next = end->next;    // (a)
            end->next = p->next;        // (b)
            p->next = end;              // (c)
            end = start->next;          // (d)
        }


class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int left, int right) {
        if(head == NULL || left == right)   return head;
        
        ListNode* dummy = new ListNode(0, head);
        
        ListNode* p = dummy;
        ListNode* start;
        ListNode* end;
        
        for(int i=1; i<left; i++){
            p = p->next;
        }
        
        start = p->next;
        end = start->next;
        for(int i=left; i<right; i++){
            start->next = end->next;    // (a)
            end->next = p->next;        // (b)
            p->next = end;              // (c)
            end = start->next;          // (d)
        }
        
        return dummy->next;
    }
};