(链表 双指针) leetcode 19. Remove Nth Node From End of List

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

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.

Follow up:

Could you do this in one pass?

 

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这个题就是移去从表尾开始的第n 个元素。emmmm，这个最好画图，便于理解。

可以用双指针。

C++代码：

/**
 * 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 *s = head;
        ListNode *e = head;
        if(!head) return NULL;
        for(int i = 0; i < n; i++)
            e = e->next;
        if(!e) return head->next;
        while(e->next){
            s = s->next;
            e = e->next;
        }
        s->next = s->next->next;
        return head;
    }
};