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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.

    题目要求我们要尽量扫描一遍,那我们就要设置两个指针了,第二个指针先前行n+1,然后第一个指针再前行,当第二个指针到达结尾时,第一个指针就是

    要删除节点的前序,同时也要考虑恰好头结点就是要删除节点的特殊情况

    /**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode* removeNthFromEnd(ListNode* head, int n) {
            ListNode *pre = head, *current = head;
            int count = 0;
            while (current)
            {
                if (!(count <= n))
                pre = pre->next;
                ++count;
                current = current->next;
            }
            if (pre == head&&count==n)
            {
                head = head->next;
                free(pre);
            }
            else
            {
                current = pre->next;
                pre->next = pre->next->next;
                free(current);
            }
            return head;
        }
    };
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  • 原文地址:https://www.cnblogs.com/csudanli/p/5883458.html
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