[LeetCode] Remove Nth Node From End of List

[Problem]
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.

[Solution]

//
// 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function

// null head
if(NULL == head || n <= 0){
return head;
}

ListNode *pre = head;
ListNode *toDel = head;
ListNode *last = head;
int count = 0;

// set the last pointer
while(last != NULL){
count++;
if(count == n){
break;
}
last = last->next;
}

// move forward
while(last->next != NULL){
last = last->next;
pre = toDel;
toDel = toDel->next;
}

// the head node is the node to be deleted
if(pre == toDel){
head = toDel->next;
}
else{
pre->next = toDel->next;
}

return head;
}
};

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