反转链表II

题目来源

反转链表II

题目描述

解答一（递归）

需要两个不同指针，一个指向第 m 个结点，另一个指向第 n 个结点。一旦有了这两个指针，我们就可以不断地交换这两个指针指向结点的数据，并将两个指针相向移动，就像字符串的情况那样。然而，链表中没有向后指针，也没有下标。因此，我们需要使用递归来 模拟 向后指针。递归中的回溯可以帮助我们模拟一个指针从第n个结点向中心移动的移动过程。

class Solution {
    
    
    private ListNode left;
    private boolean stop;
    public ListNode reverseBetween(ListNode head, int m, int n) {
    
    
        this.left=head;
        this.stop=false;
        this.recurseAndRrverse(head,m,n);
        return head;    
    }
    public void recurseAndRrverse(ListNode right,int m,int n){
    
    
        if(n==1){
    
    
            return;
        }
        right=right.next;
        if(m>1){
    
    
            this.left=this.left.next;
        }
        recurseAndRrverse(right,m-1,n-1);
        if(this.left==right||right.next==this.left){
    
    
            this.stop=true;
        }
         if(!this.stop){
    
    
            int i=this.left.val;
            this.left.val=right.val;
            right.val=i;
            this.left=this.left.next;
        }
    }
}

解答二(迭代）

class Solution {
    
    
    public ListNode reverseBetween(ListNode head, int m, int n) {
    
    
        // Empty list
        if (head == null) {
    
    
            return null;
        }

        // Move the two pointers until they reach the proper starting point
        // in the list.
        ListNode cur = head, prev = null;
        while (m > 1) {
    
    
            prev = cur;
            cur = cur.next;
            m--;
            n--;
        }

        // The two pointers that will fix the final connections.
        ListNode con = prev, tail = cur;

        // Iteratively reverse the nodes until n becomes 0.
        ListNode third = null;
        while (n > 0) {
    
    
            third = cur.next;
            cur.next = prev;
            prev = cur;
            cur = third;
            n--;
        }

        // Adjust the final connections as explained in the algorithm
        if (con != null) {
    
    
            con.next = prev;
        } else {
    
    
            head = prev;
        }

        tail.next = cur;
        return head;
    }
}