题目链接

Remove Nth Node From End of List - LeetCode

注意点

考虑删除的结点是开头的那个结点
输入给的链表是没有开头的"哑结点"的

解法

解法一：先在开头加一个哑结点，这样可以简化后面的处理。一次遍历统计有多少个结点。然后再一次遍历找到倒数第n个结点，修改next指针，结束循环。时间复杂度为O(n)

/**
 * 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) {
        ListNode* ans = new ListNode(0);
        ans->next = head;
        ListNode* p = ans;
        ListNode* q = ans->next;
        int count = 0;
        while(q != NULL)
        {
            q = q->next;
            count++;
        }
        int stop = count-n+1;
        count = 1;
        q = ans->next;
        while(q != NULL)
        {
            if(count == stop)
            {
                p->next = q->next;
                break;
            }
            p = q;
            q = q->next;
            count++;
        }
        return ans->next;
    }
};

解法二：只需要一次遍历，两个指针p和q，首先p先走n步，然后p和q一起走，当p走到结尾的时候q就走到倒数n+1个结点了。时间复杂度为O(n)

/**
 * 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) {
        ListNode* p = head;
        ListNode* q = head;
        while(n > 0)
        {
            p = p->next;
            n--;
        }
        if(p == NULL)//remove first node
        {
            return q->next;
        }
        while(p->next != NULL)
        {
            q = q->next;
            p = p->next;
        }
        q->next = q->next->next;
        return head;
    }
};

小结

解法二还是很精妙滴