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.
分析:定義兩個指針pa, pb相距爲n,其中pa從head開始。pa, pb從頭向尾行走,當pb走到末尾時,pa即爲倒數第n個數指針。
/**
* 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 *pa = head, pb = pa, pa_pre;
int dis = 0;
while(pb != NULL){
if(dis++ < n)
pb = pb->next;
else{
pa_pre = pa;
pa = pa->next;
pb = pb->next;
}
}
if(pa == head)
head = head->next;
else
pa_pre->next = pa->next;
delete pa;
return head;
}
};
有網友更簡單的程序:
//clever solution by other guy
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;
}
};