Leetcode笔记——19.删除链表倒数第n个元素

Problem

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.

Solution

• Two Pass
  这种方式比较暴力，即先遍历一边链表，然后找到链表的长度，这样我们就可以得到倒数第n个元素的正序数字。缺点就是需要遍历两次。

• One Pass
  一次遍历的话，则需要两个指针，需要一个指针先行n步，然后两个指针同时增加.(突然回忆起大二数据结构的往事，满满的回忆..记得当时的数据结构还考了99分..现在却基本都忘了)
  代码如下：

class Solution
{
public:
    ListNode* removeNthFromEnd(ListNode* head, int n)
    {
        ListNode** t1 = &head, *t2 = head;
        for(int i = 1; i < n; ++i)
        {
            t2 = t2->next;
        }
        while(t2->next != NULL)
        {
            t1 = &((*t1)->next);
            t2 = t2->next;
        }
        *t1 = (*t1)->next;
        return head;
    }
};

另一种形式

struct ListNode* front = head;
struct ListNode* behind = head;

while (front != NULL) {
    front = front->next;
    
    if (n-- < 0) behind = behind->next;
}
if (n == 0) head = head->next;
else behind->next = behind->next->next;
return head;