19. Remove Nth Node From End of List

Given the head of a linked list, remove the nth node from the end of the list and return its head.

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

Constraints:

• The number of nodes in the list is sz.
• 1 <= sz <= 30
• 0 <= Node.val <= 100
• 1 <= n <= sz

Follow up: Could you do this in one pass?

1.傻瓜办法：

/**
 * 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) {
        ListNode*curr=head;
        int count=0;
        while(curr!=NULL){
            count++;
            curr=curr->next;
        }
        int cnt=count-n+1;
        struct ListNode*dummyHead=new ListNode(0,head);
        struct ListNode*pre=dummyHead;
        count=0;
        while(pre->next!=NULL){
            count++;
            if(count==cnt){
                pre->next=pre->next->next;
            }else{
                pre=pre->next;
            }
        }
        ListNode*ret=dummyHead->next;
        delete dummyHead;
        return ret;
    }
};

注意：

我的这种方法是最容易想到的，先遍历一遍链表得到链表长度，需要注意的一点只有count++放的位置了。

2.快慢指针：

/**
 * 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) {
        ListNode*dummyHead=new ListNode(0,head);
        ListNode*fast=dummyHead;
        ListNode*slow=dummyHead;
        n++;
        while(n--)fast=fast->next;
        while(fast!=NULL){
            fast=fast->next;
            slow=slow->next;
        }
        slow->next=slow->next->next;
        ListNode*temp=dummyHead->next;
        delete dummyHead;
        return temp;
    }
};

注意：

        1.中心思想：fast比slow多走n步，但是这会导致slow最后指向的位置刚好是要删除的，所以fast应该比slow多走n+1步，保证当fast==NULL的时候，slow是要被删除的倒数第n个位置的前一个位置。（fast和slow始终距离n+1）

        2.C++代码要手动释放new的节点内存