leetcode 19. 删除链表的倒数第 N 个结点（c++）

思路分析

常规解法：双指针与栈，两种解法

题解1

/**
 * 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) {
    
    
        stack<ListNode*> stk;
        ListNode* new_head = new ListNode(0);
        int i = 0;
        while(head){
    
    
            stk.push(head);
            head = head->next;
        }
        while(!stk.empty()){
    
    
            i += 1;
            if(i == n){
    
    
                stk.pop();
                continue;
            }
            ListNode* temp = stk.top();
            stk.pop();
            ListNode* cur = new_head->next;
            new_head->next = temp;
            temp->next = cur;
        }
        return new_head->next;
    }
};

题解2

/**
 * 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 *new_head = head, *fast = head, *slow = head, *pre = NULL;
        while(n--)
            fast = fast->next;
        if(fast == NULL)
            return head->next;
        while(fast){
    
    
            fast = fast->next;
            pre = slow;
            slow = slow->next;
        }
        pre->next = slow->next;
        return head;
    }
};