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leetcode-19-remove nth node from end of list

admin 11月 12, 2020 0

题目

删除链表的倒数第n个元素

分析

首先计算出链表的长度k，删除倒数第n个元素，实际就是删除正数第k-n+1个元素

C++代码实现


 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* (ListNode* head, int n) {
        int len = get_ListNodeLength(head);
        int k = len - n + 1;
        return removeNthFromBegin(head, k);
    }

    int get_ListNodeLength(ListNode* head){
        int len = 0;
        while(head){
            len++;
            head = head->next;
        }
        return len;
    }

    ListNode* removeNthFromBegin(ListNode* head, int n){
        ListNode* pre = head;
        if(n == 1) return head->next;
        for(int i = 0; i < n - 2; i++)
            pre = pre->next;
        pre->next = pre->next->next;
        return head;
    }
};

Python代码实现

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def (self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        num = self.sumOfList(head)
        res = self.removeNthFromStart(head, num + 1 - n)
        return res
        
    def sumOfList(self,head):
        temp = head
        res = 0
        while temp is not None:
            res+=1
            temp = temp.next
        return res
    def removeNthFromStart(self, head, n):
        tmp_h = head
        tmp_b = head
        if n==1:
            return head.next
        else:
            for i in range(n-1):
                tmp_b = tmp_h
                tmp_h = tmp_h.next
            tmp_b.next = tmp_h.next
            return head