Given a linked list, remove the nthnode 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;
}
};
原文链接: https://www.cnblogs.com/kwill/p/3189736.html
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