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  • [LeetCode] 19. Remove Nth Node From End of List

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

    For 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.
    

    Note:
    Given n will always be valid.
    Try to do this in one pass.

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode* removeNthFromEnd(ListNode* head, int n) {
    12         ListNode *dummy = new ListNode(0);
    13         dummy->next = head;
    14         ListNode *ret = NULL;
    15         ListNode *prev = dummy;
    16         
    17         ListNode *p1 = head;
    18         ListNode *p2 = head;
    19         int counter = 0;
    20         
    21         while (p2){
    22             p2 = p2->next;
    23             counter++;
    24             if (counter == n){
    25                 break;
    26             }
    27         }
    28         
    29         if (counter < n){
    30             delete dummy;
    31             return head;
    32         }
    33         
    34         while(p2){
    35             p1 = p1->next;
    36             p2 = p2->next;
    37             prev = prev->next;
    38         }
    39         
    40         prev->next = p1->next;
    41         
    42         ret = dummy->next;
    43         delete dummy;
    44         return ret;
    45     }
    46 };
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  • 原文地址:https://www.cnblogs.com/amadis/p/5926516.html
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