问题描述:
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?
题源:here;代码实现:here
思路:
两个方案,一个简单的把每一个中间节点都存下来;一个是巧妙的利用中间变量来变相记录链表长度。
方案1
ListNode* removeNthFromEnd_1(ListNode* head, int n) {
int idx = 0;
vector<ListNode*> nodes;
ListNode* temp = head;
while (temp){
nodes.push_back(temp);
temp = temp->next;
}
int list_len = nodes.size();
if (list_len == 1) return NULL;
else if (list_len == 2){
if (n == 1){
delete head->next;
head->next = NULL;
return head;
}
else{
ListNode* temp = head;
head = head->next;
delete temp;
return head;
}
}
if (list_len == n){
ListNode* temp = head;
head = head->next;
delete temp;
return head;
}
ListNode* q = nodes[list_len-n-1];
ListNode* p = q->next;
q->next = p->next;
delete p;
return head;
}
方案2
方案二使用了两个指针,第一个先沿着输入的链表前进n步,第二个指向链表头;然后让两个指针同时向后移动,直到第一个指针指向链表的尾部,如下图所示(图片来源:here):
代码如下:
ListNode* removeNthFromEnd(ListNode* head, int n){
ListNode *feakhead;
feakhead = new ListNode(-1);
feakhead->next = head;
head = feakhead;
ListNode *p1, *p2;
p1 = p2 = head;
for (int i = 0; i < n; i++){
p1 = p1->next;
}
while (p1->next){
p1 = p1->next;
p2 = p2->next;
}
ListNode* temp;
temp = p2->next;
p2->next = p2->next->next;
delete temp;
return head->next;
}