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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.struct ListNode {
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
ListNode *pTemp = head;
ListNode *hash[1000] = {nullptr};
int len = 0;
while (pTemp) {
hash[len] = pTemp;
pTemp = pTemp->next;
++len;
}
int tar = len - n;
delete(hash[tar]);
if (tar == 0) {
head = hash[tar + 1];
} else {
hash[tar - 1]->next = hash[tar + 1];
}
return head;
}
};LeetCode之19---Remove Nth Node From End of List
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原文地址：http://blog.csdn.net/jung_zhang/article/details/51167653
