[leetcode]19 删除链表的倒数第 N 个结点 | 链表模拟
题目链接:https://leetcode.cn/problems/remove-nth-node-from-end-of-list/
该链表中,head节点对应有值
要想知道倒数第几个节点对应正序哪个节点,需要先进行遍历知道整个链表的长度
倒数第n个元素就是第len - n + 1个元素
然后让前一个结点指向当前节点的下一个节点便完成任务
code:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
int cnt = 0;
ListNode* h2 = head;
while(head != nullptr) {
head = head->next;
cnt ++;
}
// std::cout << cnt << std::endl;
int pos = cnt - n + 1;
if(pos == 1) return h2->next;
ListNode* pre = nullptr;
head = h2;
for(int i = 1;i <= cnt;i ++) {
if(i == pos) {
/**
// 1
if(head->next == nullptr) pre->next = nullptr;
else {
pre->next = head->next;
}
**/
// 2
pre->next = head->next;
}
pre = head;
head = head->next;
}
head = h2;
return head;
}
};