始于
分类：LeetCode
Tags: [ LeetCode ]
LeetCode 百题计划 - 19. Remove Nth Node From End of List
LeetCode 19. Remove Nth Node From End of List
19. Remove Nth Node From End of List
Difficulty: medium
Given a linked list, remove the n-th node from the end of list and return its head.
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.
Follow up:
Could you do this in one pass?
1st Version:
/**
* 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) {
vector<ListNode*> nodes;
auto node = head;
while (node != NULL) {
nodes.push_back(node);
node = node->next;
}
int po = nodes.size() - n;
if (po - 1 < 0) {
delete nodes[po];
if (po + 1 <= nodes.size() - 1)
head = nodes[po + 1];
else
head = NULL;
} else {
delete nodes[po];
if (po + 1 <= nodes.size() - 1)
nodes[po - 1]->next = nodes[po + 1];
else
nodes[po - 1]->next = NULL;
}
return head;
}
};
Rate:
Runtime: 4 ms, faster than 86.11% of C++ online submissions for Remove Nth Node From End of List.
Memory Usage: 8.8 MB, less than 6.58% of C++ online submissions for Remove Nth Node From End of List.
真是搞不懂我是什么弱智, 为什么要这么写, 还搞一个vector来存, 这样的话直接遍历两遍不就好了.
重写了一下, 时间和空间其实都已经达到预期了:
2nd Version:
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {
size_t len = 0u;
auto itr = head;
while (itr != NULL) {
itr = itr->next;
++len;
}
int po = len - n;
itr = head;
if (po == 0) {
head = head->next;
delete itr;
} else {
for (int i = 0; i < po - 1; ++i) {
itr = itr->next;
}
auto newNext = itr->next->next;
delete itr->next;
itr->next = newNext;
}
return head;
}
};
Rate:
Runtime: 4 ms, faster than 86.11% of C++ online submissions for Remove Nth Node From End of List.
Memory Usage: 8.4 MB, less than 98.68% of C++ online submissions for Remove Nth Node From End of List.
但是题目中叫我们挑战一遍拿到倒数第n个元素, 没经验, 想不出来.
3rd Version (Java):
// https://leetcode.com/problems/remove-nth-node-from-end-of-list/discuss/359826/Java-recursive-solution%3A-Runtime%3A-0-ms-faster-than-100.00-O(n)
class Solution {
int cnt = 0;
public int findN(ListNode current, int n) {
if(current == null) {
return (n + 1);
}
n = findN(current.next, n);
if(n != -1)
n--;
if(n == 0) {
current.next = current.next.next;
n = -1;
}
return n;
}
public ListNode removeNthFromEnd(ListNode head, int n) {
int isDone = findN(head, n);
if(isDone != -1) {
head = head.next;
return head;
}
return head;
}
}
奇妙的方法, 递归. 不多解释了.