[Leetcode][Python]19: Remove Nth Node From End of List
# -*- coding: utf8 -*-'''
__author__ = 'dabay.wang@gmail.com'
19: Remove Nth Node From End of List
https://oj.leetcode.com/problems/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.
===Comments by Dabay===
那种两个指针同时走的解法，应该是不符合题意的。人家要求的一次pass，你两个指针同时走，实际上是2次pass了。
思路一：
遍历的时候，把每个node放到一个栈中，然后更加n弹出到相应的位置删除节点。空间复杂度为ListNode的长度。
思路二：
用一个大小为n+1的队列来记录指针之前的n个节点。当指针到最后的时候，删除队列中的第二元素。空间复杂度为n+1。
'''
# Definition for singly-linked list.
class ListNode:
def __init__(self, x):
self.val = x
self.next = None
class Solution:
# @return a ListNode
def removeNthFromEnd(self, head, n):
cursor = new_head = ListNode(0)
new_head.next = head
queue = []
while cursor:
queue.append(cursor)
if len(queue) > n + 1:
queue.pop(0)
cursor = cursor.next
previous = queue.pop(0)
to_del = queue.pop(0)
previous.next = to_del.next
return new_head.next
# cursor = new_head = ListNode(0)
# new_head.next = head
# stack = []
# while cursor:
# stack.append(cursor)
# cursor = cursor.next
# while n > 1:
# stack.pop()
# n = n - 1
# to_del = stack.pop()
# previous = stack.pop()
# previous.next = to_del.next
# return new_head.next
def main():
sol = Solution()
root = ListNode(1)
n2 = ListNode(2)
n3 = ListNode(3)
n4 = ListNode(4)
root.next = n2
n2.next = n3
n3.next = n4
sol.removeNthFromEnd(root, 1)
while root:
print "%s -> " % root.val,
root = root.next
print " End"
if __name__ == '__main__':
import time
start = time.clock()
main()
print "%s sec" % (time.clock() - start)
页:
[1]