索引：[LeetCode] Leetcode 题解索引 (C/Java/Python/Sql)Github: https://github.com/illuz/leetcode019.Remove_Nth_Node_From_End_of_List (Easy)链接：题目：https://oj.le
索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/sql)
Github:
https://github.com/illuz/leetcode
019.Remove_Nth_Node_From_End_of_List (Easy)
链接：
题目：https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
代码(github)：https://github.com/illuz/leetcode
题意：
删除1个单向链表的倒数第 N 个节点。
分析：
- 直接摹拟，先算出节点数，再找到节点删除
- 用两个指针，1个先走 N 步，然后再1起走。
这里用 C++ 实现第1种， 用 Python 实现第2种。
Java 的话和 C++/Python 差不多，不写出来了。
代码：
C++:
class Solution {
public:
ListNode *removeNthFromEnd(ListNode *head,int n) {
if (n == 0)
return head;
// count the node number
int num = 0;
ListNode *cur = head;
while (cur != NULL) {
cur = cur->next;
num++;
}
if (num == n) {
// remove first node
ListNode *ret = head->next;
delete head;
return ret;
} else {
// remove (cnt-n)th node
int m = num - n - 1;
cur = head;
while (m--)
cur = cur->next;
ListNode *rem = cur->next;
cur->next = cur->next->next;
delete rem;
return head;
}
}
};
Python:
class Solution:
# @return a ListNode
def removeNthFromEnd(self,head,n):
dummy = ListNode(0)
dummy.next = head
p,q = dummy,dummy
# first 'q' go n step
for i in range(n):
q = q.next
# q & p
while q.next:
p = p.next
q = q.next
rec = p.next
p.next = rec.next
del rec
return dummy.next
服务热线 19970861797
