[LeetCode] 19. Remove Nth Node From End of List

[grandyang]

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Given the head of a linked list, remove the nth node from the end of the list and return its head.

Example 1:

**Input:** head = [1,2,3,4,5], n = 2
**Output:** [1,2,3,5]

Example 2:

**Input:** head = [1], n = 1
**Output:** []

Example 3:

**Input:** head = [1,2], n = 1
**Output:** [1]

Constraints:

• The number of nodes in the list is sz.
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

Follow up: Could you do this in one pass?

这道题让我们移除链表倒数第N个节点，限定n一定是有效的，即n不会大于链表中的元素总数。还有题目要求一次遍历解决问题，那么就得想些比较巧妙的方法了。比如首先要考虑的时，如何找到倒数第N个节点，由于只允许一次遍历，所以不能用一次完整的遍历来统计链表中元素的个数，而是遍历到对应位置就应该移除了。那么就需要用两个指针来帮助解题，pre 和 cur 指针。首先 cur 指针先向前走N步，如果此时 cur 指向空，说明N为链表的长度，则需要移除的为首元素，那么此时返回 head->next 即可，如果 cur 存在，再继续往下走，此时 pre 指针也跟着走，直到 cur 为最后一个元素时停止，此时 pre 指向要移除元素的前一个元素，再修改指针跳过需要移除的元素即可，参见代码如下：

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *pre = head, *cur = head;
        for (int i = 0; i < n; ++i) cur = cur->next;
        if (!cur) return head->next;
        while (cur->next) {
            cur = cur->next;
            pre = pre->next;
        }
        pre->next = pre->next->next;
        return head;
    }
};

Github 同步地址：

#19

类似题目：

Linked List Cycle

Linked List Cycle II

Swapping Nodes in a Linked List

Delete N Nodes After M Nodes of a Linked List

Delete the Middle Node of a Linked List

参考资料：

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/8812/My-short-C%2B%2B-solution

https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/8804/Simple-Java-solution-in-one-pass

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