[Leetcode][Python]19: Remove Nth Node From End of List

# -*- coding: utf8 -*-‘‘‘__author__ = ‘[email protected]‘

19: Remove Nth Node From End of Listhttps://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)