题目
Given a linked list, remove the nth node from the end of list and return its head.
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
先从头遍历n个结点，找到需要删除的前一个结点removePre，然后每次移动一次游标，移动一次removePre。注意如果删除的是head，需要做处理，因为removePre此时应该没有值。
代码
/**
* 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*
next = head;
ListNode* removePre =
NULL;
int firstN = n;
while (n >
0) {
next =
next->
next;
n--;
}
if (
next!=
NULL) {
removePre = head;
}
else {
return head->
next;
}
while (
next->
next !=
NULL) {
next =
next->
next;
removePre = removePre->
next;
}
ListNode* tmp = removePre->
next;
removePre->
next = (tmp==
NULL)?tmp:tmp->
next;
return head;
}
};
