标签：style blog class code c java
题意: 给出单链表头指针head和整数n, 要求删除链表的倒数第n个结点, 这里n保证是合法的输入.
我的思路....其实我没大明白题目建议的one pass是什么意思, 可能就是遍历一遍链表的, 不过我还是秉着能A掉就万岁的心态...我还是首先记录了链表的长度, 然后删除第len - n + 1个结点
附上代码:
/**
* 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) {
if (head == NULL or (head->next == NULL and n == 1)) {
return head = NULL;
}
int len = 0, cnt = 1;
ListNode *p = head;
// count the length of the list
while (p != NULL) {
len++;
p = p->next;
}
if (n == len) {
return head = head->next;
}
p = head;
// find the (len-n)th node from the head
while (p != NULL and cnt < len-n) {
cnt++;
p = p->next;
}
p->next = p->next->next;
return head;
}
};
后来看了discuss, 看到老外们讨论的one pass...说真的, 我没想出来这种思路. 使用两个指针, fast和slow, fast比last多走n个结点, 当fast走到链表尾部的时候, slow就是要删除的结点的前一个结点. 可是这样还是遍历了两遍啊,( 除了最后n个结点. )
附上代码:
1 /**
2 * Definition for singly-linked list.
3 * struct ListNode {
4 * int val;
5 * ListNode *next;
6 * ListNode(int x) : val(x), next(NULL) {}
7 * };
8 */
9 class Solution {
10 public:
11 ListNode *removeNthFromEnd(ListNode *head, int n) {
12 if (head == NULL) {
13 return head;
14 }
15 ListNode *fast = head, *slow = head;
16 int tmp_n = n;
17 // fast pointer to the Nth node
18 while (tmp_n--) {
19 fast = fast->next;
20 }
21 if (fast == NULL) {
22 return head = head->next;
23 }
24 // move fast and slow simulataneously
25 while (fast != NULL and fast->next != NULL) {
26 fast = fast->next;
27 slow = slow->next;
28 }
29 slow->next = slow->next->next;
30 return head;
31 }
32 };
LeetCode---Remove Nth Node From End of List,布布扣,bubuko.com
LeetCode---Remove Nth Node From End of List
标签：style blog class code c java
原文地址：http://www.cnblogs.com/Stomach-ache/p/3726274.html
