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.
class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { // Start typing your C/C++ solution below // DO NOT write int main() function if (!head || n <= 0) return head; ListNode dummynode(0), *h = &dummynode,*p = head,*e = p; h->next = head; int k = 0; for(;k < n && e; k++){ e = e->next; } //n正好为总长度 if (!e){ return head->next; } while(p && e && e->next){ p = p->next; e = e->next; } if (p) p->next = p->next->next; return h->next; } };