給定一個鏈表,移除從尾部起的第 n 個結點,并且返回它的頭結點。
例如,給定鏈表:1->2->3->4->5,n = 2。
在移除尾部起第二個結點后,鏈表將變成:1->2->3->5。
備注:
給定的 n 是有效的,代碼盡量一次通過。
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.
class Solution{
public:
ListNode* removeNthFromEnd(ListNode* head, int n){
ListNode newHead(0);
newHead.next = head;
count = n;
return solution1(&newHead);
}
private:
int count;
ListNode* solution1(ListNode* newHead){
subSol1(newHead);
return newHead->next;
}
bool subSol1(ListNode* node){
if(!node) return false;
if(subSol1(node->next)) return true;
if(count--) return false;
ListNode *tmp = node->next;
node->next = tmp->next;
delete tmp;
return true;
}
};