[LeetCode_19] Remove Nth Node From End of List_删除链表的倒数第N个节点
leetcode_19
题目描述:
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 becomes1->2->3->5.
思路:
定义两个指针,一个叫做slow,一个叫做 fast,fast就是跑在前面的,slow就是跑在后面的。你要找倒数第n(n是合理的输入,不会超过链表长度)个,那我先让fast从链表头往前跑n步,当fast站在第n个节点这个位置的时候,slow开始从起点跑,fast也同步从n跑,当fast到终点的时候,slow所在位置的下一个节点就是倒数第n个节点了。
代码:
方法1:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
struct ListNode ret, *p = &ret, *q = head;
ret.next = head;
while(n--) {
q = q -> next;
}
while(q) {
p = p->next;
q = q->next;
}
q = p->next;
p->next = q->next;
free(q);
return ret.next;
}
方法2:
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {
int len = 0;
struct ListNode ret, *p = head, *q;
ret.next = head;
while(p) {
p = p->next;
len++;
}
p = &ret;
len -= n;
while(len--) {
p = p->next;
}
q = p->next;
p->next = q->next;
free(q);
return ret.next;
}