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Leetcode: Remove Nth Node From End of List
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：two pointers.
class Solution {public: ListNode *removeNthFromEnd(ListNode *head, int n) { if(head == NULL) return head; ListNode *dummy = new ListNode(-1); dummy->next = head;
//move fast pointer to (n+1)th node from the head ListNode *fast = head; for(int i = 0; i < n; i++) fast = fast->next; //move fast to the end of the list ListNode *prev = dummy, *p = head; while(fast){ fast = fast->next; prev = p; p = p->next; } prev->next = p->next; return dummy->next; }};
Leetcode: Remove Nth Node From End of List
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