[Medium] 92. Reverse Linked List II

#문제 설명

Leetcode 92번 문제는 1부터 n까지 정렬된 Linked List와 left와right 값이 주어진다. 그럼 left부터right까지의 노드들을 reverse하면 된다. 

 

#문제 풀이

        List<ListNode> list = new ArrayList<>();
        ListNode curr = head;
        
        while (curr != null) {
            list.add(curr);
            curr = curr.next;
        }
        
        Object[] arr = list.toArray();
            
        left = left - 1;
        right = right - 1;
        
        while (left < right) {
            ListNode temp = (ListNode) arr[left];
            arr[left] = arr[right];
            arr[right] = (ListNode) temp;
            left++;
            right--;
        }
        
        ListNode sentinel = new ListNode(-1);
        curr = sentinel;
        
        for (int i = 0; i < arr.length; i++) {
            curr.next = (ListNode) arr[i];
            curr = curr.next;
        }
        
        curr.next = null;
        return sentinel.next;

Array를 사용해 풀수있다.

 

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;
    
}

Dummy Node를 사용해 one pass로 풀수있다.