92. Reverse Linked List II

链表区间反转

最新推荐文章于 2021-09-25 16:26:57 发布

原创最新推荐文章于 2021-09-25 16:26:57 发布 · 131 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法

数据结构&算法专栏收录该内容

94 篇文章

订阅专栏

本文介绍了一种在给定范围内反转链表的有效方法。通过两次遍历实现：第一次找到待反转链表部分的前一个节点；第二次进行就地反转操作。提供了两种实现方案，一种在反转结束后链接链表，另一种在每次反转时即进行链接。

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:

Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:

Given m, n satisfy the following condition:

1 ≤ m ≤ n ≤ length of list.

比较简单一次ac 4ms

想法也很传统找到需要反转的起始点m 在n之前一直反转然后m-1->n m->n+1

代码如下

    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null) return head;
        
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        head = dummy;
        int i = 1;
        while (i < m) {
            head = head.next;
            i++;
        }
        ListNode pre = head.next;
        if (pre  == null) return head;
        
        ListNode cur = pre.next;
        while (cur!=null && i<n) {
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
            i++;
        }
        
        head.next.next = cur;
        head.next = pre;
        return dummy.next;
    }

while的地方可以用for循环简化一下

下面的solution只要三毫秒想法有点特别

我是等反转结束再链接 solution是每次反转都链接

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;
    
}