Code & Func

打卡，第14天

今天的题目是Remove Nth Node From End of List,一开始以为是道很简单的题目，后来看dicuss时才发现是自己没看清题目。

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.

一开始没看到Try to do this in one pass.,然后就用两遍遍历方法去做了:

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int getRightN(ListNode *head,int n) {
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    ListNode *p = head;
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    int size = 0;
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    while(p != nullptr) {
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        p = p->next;
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        size++;
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    }
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    return size - n;
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}
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ListNode* removeNthFromEnd(ListNode* head, int n) {
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    ListNode h(0);
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    h.next = head;
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    ListNode *p = &h;
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    int k = getRightN(p,n);
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    while(--k)
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        p=p->next;
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    head = p->next;
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    p->next = head->next;
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    delete head;
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    return h.next;
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}

上面这个方法太简单了，还是看看在dicuss中的方法吧：

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ListNode *removeNthFromEnd(ListNode *head, int n)
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{
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    if (!head)
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        return nullptr;
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    ListNode new_head(-1);
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    new_head.next = head;
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    ListNode *slow = &new_head, *fast = &new_head;
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    for (int i = 0; i < n; i++)
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        fast = fast->next;
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    while (fast->next)
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    {
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        fast = fast->next;
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        slow = slow->next;
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    }
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    ListNode *to_de_deleted = slow->next;
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    slow->next = slow->next->next;
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    delete to_be_deleted;
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    return new_head.next;
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}

这个看起来会比较简单，想法就是利用快慢指针去做，先让fast指针先走n步，然后在fast指针和slow指针一起移动，这样fast和slow始终保持着n个节点的距离，当fast为最后一个节点时，slow就指向倒数第n+1个节点，这时就可以把倒数第n个节点删掉了。

有一个更简洁的版本，不过有点难懂就是了：

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

这个和上一个的思路其实是完全一样的，只是实现方法思路不一样就是，这里的t1是指向某个节点(包括一开始时虚拟的头结点）的next指针的指针。