O(n)时间复杂度

题目链接

程序

/**
 * Definition of singly-linked-list:
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param head: The first node of linked list.
     * @param n: An integer
     * @return: The head of linked list.
     */
    ListNode * removeNthFromEnd(ListNode * head, int n) {
        // write your code here
        if(head == NULL)    return head;
        int len = 0;
        for(ListNode *node = head; node != NULL; node = node->next)
            len++;
        if(len == n){//如果删除的为头结点
            return head->next;//直接返回头结点后面的链表
        }
        ListNode *fast = head;
        for(int i = 0; i <= n; i++)
            fast = fast->next;
        ListNode *slow = head;
        while(fast != NULL){//通过快慢指针找到倒数的节点
            fast = fast->next;
            slow = slow->next;
        }
        ListNode *tmp = slow->next;
        slow->next = slow->next->next;
        delete tmp;
        return head;
    }
};