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.

題目鏈接

思路
只遍歷一遍，這裏就想到用兩個指針，第一個先走n步，然後再一起遍歷，第一個到達尾部的時候，第二個指向的剛好是倒數第n個，然後把它刪掉即可。

代碼（C）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n) 
{
    struct ListNode *newhead = (struct ListNode *)malloc(sizeof(struct ListNode));
    if (NULL == newhead)
        return head;
    newhead->next = head;

    struct ListNode *fast = newhead;
    struct ListNode *pre = NULL;

    int i = 0;
    while(i < n)
    {
        fast = fast->next;
        i++;
    }
    pre = newhead;
    while(fast->next != NULL)
    {
        pre = pre->next;
        fast = fast->next;
    }
    pre->next = pre->next->next;

    return newhead->next;
}

這裏定義一個newhead的原因是：當n等於鏈表長度時，需要刪除頭指針，這時候head要被刪除。

代碼（python）

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        newhead = ListNode(0)
        newhead.next = head
        fast = head
        index = 0
        while index < n:
            fast = fast.next
            index += 1
        pre = newhead
        while fast:
            pre = pre.next
            fast = fast.next
        pre.next = pre.next.next
        return newhead.next