删除链表第n个节点

最新推荐文章于 2024-10-22 10:12:23 发布

原创最新推荐文章于 2024-10-22 10:12:23 发布 · 514 阅读

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本文介绍了一种使用双指针技术在单次遍历中删除链表倒数第N个节点的方法。通过保持两个指针之间的固定距离(n-1)，当后指针到达链表末尾时，前指针恰好指向待删除节点的前一个位置。

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Given a linked list, remove the n^th 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.

依然是双指针思想，两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;
            
        ListNode *pPre = NULL;
        ListNode *p = head;
        ListNode *q = head;
        for(int i = 0; i < n - 1; i++)
            q = q->next;
            
        while(q->next)
        {
            pPre = p;
            p = p->next;
            q = q->next;
        }
        
        if (pPre == NULL)
        {
            head = p->next;
            delete p;
        }
        else
        {
            pPre->next = p->next;
            delete p;
        }
        
        return head;
    }
};