問題描述:
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.
思路:
題目看來挺簡單,主要就是設置一個前行節點一個後行節點,同步控制兩節點距離一直到鏈表末端。但是實踐完成代碼的思路上還是有一點點技巧和注意的。
這裏我用了一個鏈表附加頭節點,這樣在刪除頭結點問題上可以很方便解決;然後注意要得到的節點應該是所要求刪除節點的前一個節點;還有就是最後返回鏈表時,要返回更新後的head節點,因爲head可能有改變。
代碼:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode* prehead = new ListNode(-1);
prehead->next = head;
ListNode* bias = prehead;
ListNode* targetpre = prehead;
//bias move forward n steps
for(int i = 0; i < n; i++)
{
if(bias == NULL) return head;
bias = bias->next;
}
//bias point to the end, at the same time, target point to the previous node of the nth from the end.
while(bias->next != NULL)
{
bias = bias->next;
targetpre = targetpre->next;
}
ListNode* temp = targetpre->next;
targetpre->next = temp->next;
head = prehead->next; //update the new head
delete temp;
delete prehead;
return head;
}
};