删除链表的倒数第N个节点给定一个链表, 删除链表的倒数第 n 个节点, 并且返回链表的头结点. <div class="s

解题思路快慢指针, 遍历找出节点, 跳过. 代码 <di

LeetCode题解解题思路根据栈先进后出的原则, 全部进

LeetCode题解解题思路快慢指针, 遍历找出节点, 跳过. <

删除链表的倒数第N个节点 about pisecespeng.record.me HOT 3 CLOSED

PisecesPeng commented on September 25, 2024

删除链表的倒数第N个节点

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Comments (3)

PisecesPeng commented on September 25, 2024

解题思路

快慢指针, 遍历找出节点, 跳过.

代码

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}

private static ListNode func(ListNode head, int n) {
   ListNode newHead = new ListNode(0);
   newHead.next = head;
   // 双指针法
   ListNode head1 = newHead;
   ListNode head2 = newHead;
   // head1 提前先走n步
   while ((n--) > 0) {
       head1 = head1.next;
   }
   // head1到链表最后, head1、head2之间间隔n
   while (head1.next != null) {
       head1 = head1.next;
       head2 = head2.next;
   }

   if (head1 != head2)
       head2.next = head2.next.next;

   return newHead.next;
}

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PisecesPeng commented on September 25, 2024

LeetCode题解

解题思路

根据栈先进后出的原则,
全部进栈后, 弹出的第n个节点就是需要删除的节点,
得到其前节点, 再跳过待删除节点即可.

代码

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}

public ListNode func(ListNode head, int n) {
    ListNode dummy = new ListNode(0, head);
    Deque<ListNode> stack = new LinkedList<ListNode>();
    ListNode cur = dummy;
    while (cur != null) {
        stack.push(cur);
        cur = cur.next;
    }
    for (int i = 0; i < n; ++i) {
        stack.pop();
    }
    ListNode prev = stack.peek();
    prev.next = prev.next.next;
    ListNode ans = dummy.next;
    return ans;
}

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PisecesPeng commented on September 25, 2024

LeetCode题解

解题思路

快慢指针, 遍历找出节点, 跳过.

代码

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
    }
}

public ListNode func(ListNode head, int n) {
    ListNode dummy = new ListNode(0, head);
    ListNode first = head;
    ListNode second = dummy;
    for (int i = 0; i < n; ++i) {
        first = first.next;
    }
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    second.next = second.next.next;
    ListNode ans = dummy.next;
    return ans;
}