题目链接
注意点
解法
小结

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

转载于:https://www.cnblogs.com/multhree/p/10345974.html

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Remove Nth Node From End of List - LeetCode

题目链接

注意点

解法

小结

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