leetcode-华为专题-19. 删除链表的倒数第 N 个结点

 

 

/**
 * 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* H = new ListNode();  // 加一个头节点，避免了对只有一个节点情况的判断
        H->next = head;
        int len = getlen(H);
        if(head==0)
            return NULL;
        int cnt = len - n-1;
        ListNode* p = H;

        while(cnt--){
            p = p->next;
        }
        // cout<<"val: "<<p->val<<endl;
        p->next = p->next->next;
        return H->next;

    }
    int getlen(ListNode* head){
        if(head==NULL)
            return 0;
        int count = 0;
        while(head){
            count++;
            head = head->next;
        }
        return count;
    }
};