题目描述
Given a linked list, remove the n-th node from the end of list and return its head.
Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?
Example:
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
解题思路
- 比如首先要考虑的时，如何找到倒数第N个节点，由于只允许一次遍历，所以不能用一次完整的遍历来统计链表中元素的个数，而是遍历到对应位置就应该移除了。
- 那么就需要用两个指针来帮助解题，pre 和 cur 指针。
- 首先 cur 指针先向前走N步，如果此时 cur 指向空，说明N为链表的长度，则需要移除的为首元素，那么此时返回 head->next 即可，如果 cur 存在，再继续往下走，此时 pre 指针也跟着走，直到 cur 为最后一个元素时停止，此时 pre 指向要移除元素的前一个元素，再修改指针跳过需要移除的元素即可
C++代码
/**
* 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) {
if(head == nullptr){
return head;
}
ListNode* fast = head;
ListNode* slow = head;
while(n > 0){
fast = fast -> next;
n--;
}
// 重点 检查 这里是否为空
if(fast == nullptr){
return head -> next;
}
while(fast -> next){
fast = fast -> next;
slow = slow -> next;
}
slow -> next = slow -> next -> next;
return head;
}
};
/**
* 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) {
if(head == NULL){
return NULL;
}
ListNode* pre = head, *cur = head;
for(int i = 0; i < n; i++){
cur = cur -> next;
}
if(cur == NULL){
return head -> next;
}
while(pre && cur -> next){
pre = pre -> next;
cur = cur -> next;
}
pre -> next = pre -> next -> next;
return head;
}
};
