Remove Nth Node From End of List
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.
分析:
设两个指针p,q,让q先走n步,然后p和q一起走,直到q走到尾节点,删除p->next即可。
代码:
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head, int n) {
ListNode dummy(-1);
dummy.next = head;
ListNode *p = &dummy, *q = &dummy;
for (int i = 0; i < n; i++) { //q先走n步
q = q->next;
}
while (q->next) { //一起走
p = p->next;
q = q->next;
}
ListNode *tmp = p->next;
p->next = p->next->next;
delete tmp;
return dummy.next;
}
};