[leetcode] 19. Remove Nth Node From End of List python實現【easy】

1. Remove Nth Node From End of List My Submissions QuestionEditorial Solution

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.

意思是說給你一個單鏈表， 然後讓你刪掉從後往前的第n個，比如這個1−>2−>3−>4−>5 ,刪掉倒數第二個就變成了1−>2−>3−>5.
而且有要求，最好跑一次。。
那想法就是記錄一下唄，每次跑的時候都記一下它前第n個位置那個節點。如果當前節點變成了空（也就是到了結尾），則刪掉對應的記錄的那個節點（前第n個節點）。

雙指針的意思

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if head == None:
            return
        p = q = head
        r = None
        i = 0
        while q != None:
            if i != n:
                i +=1
                q = q.next
                continue
            else:
                r = p
                p = p.next
                q = q.next
        if r ==None:
            return head.next
        r.next = p.next
        return head