[leetcode] 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.

tag: linked-list, two pointers

用两个指针(快指针和慢指针)。

快指针先往前走n步,然后两个指针一起走,直到快指针到链表尾。

要注意n=length of list 的情况,也即要删掉链表里的第一个元素;还有要确定删掉链表中的某个元素时,实际上你需要将指针停在该元素的前一个元素,再进行删除。当然你也可以停到要删除的那个元素,再把下一个元素的value拷贝到该元素中,删除下一个元素,当然这是后话。

/**
 * 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 * p, * q;
        p = q = head;
        while(n--){
            p = p->next;
        }
        if(p == NULL){
            return head->next;
        }
        while(p->next != NULL){
            p = p->next;
            q = q->next;
        }
        q->next = q->next->next;
        return head;
    }
};

Untitled

 

 

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