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LeetCode[Linked List]: 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.
这个题目我想到两种解法。时间复杂度均为O(N),空间复杂度有所区别,一种是O(N),一种是O(1)。
另外,这个题目我觉得需要注意的地方是应该delete掉删除的节点,防止内存泄露。
空间复杂度为 O(N) 的解法
需要用到一个vector保存所有节点的地址,具体代码如下:
C++ code ListNode *removeNthFromEnd(ListNode *head, int n) {
vector<ListNode *> list;
for (ListNode *iter = head; iter != NULL; iter = iter->next)
list.push_back (iter);
if (n == list.size()) {
head = list[0]->next;
delete list[0];
return head;
}
list[list.size() - n - 1]->next = list[list.size() - n]->next;
delete list[list.size() - n];
return list[0];
}
空间复杂度为 O(1) 的解法
设置两个指针,fast较low快n个节点。具体代码如下:
C++ code ListNode *removeNthFromEnd(ListNode *head, int n) {
ListNode *fast = head, *low = head, *delNode;
for (int i = 0; i < n; ++i)
fast = fast->next;
if (!fast) {
delNode = head;
head = head->next;
delete delNode;
return head;
}
while (fast->next) {
fast = fast->next;
low = low ->next;
}
delNode = low->next;
low->next = low->next->next;
delete delNode;
return head;
}
LeetCode[Linked List]: Remove Nth Node From End of List
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