leetcode 19 刪除鏈表尾部第N個元素（Remove Nth Node From End of List）

題目描述

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

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

思路

使用兩個指針left，right

• right向後移動，直到right - left = n - 1
• left/right一起往後移動，直到鏈表尾部
• 刪除left指向的元素

代碼

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head == NULL) return NULL;
        
        ListNode *left = head;
        ListNode *leftPrev = head;
        ListNode *right = head;
        while (right->next != NULL) {
            right = right->next;
            if (n > 1) {
                --n;
                continue;
            }
            leftPrev = left;
            left = left->next;
        }
        
        if (left == head) {
            return left->next;
        }
        leftPrev->next = left->next;
        return head;
    }
};

執行效率

Runtime: 4 ms, faster than 85.64% of C++ online submissions for Remove Nth Node From End of List.
Memory Usage: 8.6 MB, less than 69.74% of C++ online submissions for Remove Nth Node From End of List.