Special list: N penultimate node to delete the list Leetcode 19 + Leetcode 24 twenty-two switching node in the linked list chain rotation + Leetcode 61
Reciprocal Leetcode 19 to delete the list of N nodes
Title Description
Given a list, delete list reciprocal of n nodes, and returns to the head node list.
Example:
给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后，链表变为 1->2->3->5.
Description:
N is given to ensure effective.
Advanced:
You can try to use one pass to achieve it?
answer
As to excitement, then we carry out in the end: to directly address this advanced request.
We need two nodes: == a departure from the head node; a node from scratch starting position + N nodes. == then when the second node reaches the end of time, under the head node of a node is to delete the location. Then add delete position is the point of discussing the border can easily solve this question:
There are a border point these types of situations:
- Node is removed when the head node
- Deleted node is the end node
Entire function like this:
class Solution(object):
def removeNthFromEnd(self, head, n):
front = head
second = None
count = 0
flag = 0
while front:
if front.next:
front = front.next
count = count + 1
if count > n-1:
if flag == 0:
second = head
flag = 1
else:
second = second.next
else:
break
if second == None:
res = head.next
head = None
return res
else:
if second.next:
temp = second.next.next
second.next = temp
temp = None
else:
second.next = None
return head
operation result
When run with: 16 ms
Memory consumption: 11.7 MB
Leetcode 24 twenty-two switching node in the linked list
Title Description
Given a list pairwise exchange with adjacent nodes, and return the list to the exchange.
You can not just simply change the internal node values, but need to be the actual node exchange.
Example:
给定 1->2->3->4, 你应该返回 2->1->4->3.
answer
Not any particular method is to traverse back from the beginning again, between two to exchange, if no successor node after only one node or exit traversal. The whole process is as follows:
class Solution(object):
def swapPairs(self, head):
if not head:
return head
cur1 = head
cur2 = head.next
if not cur2:
return head
while True:
if cur1 == head:
cur1.next = cur2.next
cur2.next = cur1
head = cur2
else:
if not cur1.next:
break
else:
if not cur1.next.next:
break
else:
temp1 = cur1.next
temp2 = cur1.next.next
temp1.next = temp2.next
cur1.next = temp2
temp2.next = temp1
cur1 = temp1
return head
operation result
Leetcode 61 rotating chain
Title Description
Given a list, the list of rotation, each of the node list to the right by k positions, wherein k is non-negative.
Example 1:
输入: 1->2->3->4->5->NULL, k = 2
输出: 4->5->1->2->3->NULL
解释:
向右旋转 1 步: 5->1->2->3->4->NULL
向右旋转 2 步: 4->5->1->2->3->NULL
Example 2:
输入: 0->1->2->NULL, k = 4
输出: 2->0->1->NULL
解释:
向右旋转 1 步: 2->0->1->NULL
向右旋转 2 步: 1->2->0->NULL
向右旋转 3 步: 0->1->2->NULL
向右旋转 4 步: 2->0->1->NULL
answer
In fact, not so obedient, he said a few times to move move several times, if we can find the length of the entire list, just move on it.
First, the first connected end to end of the list, together into a ring (when the connection chain length can be determined), and then the head node starts walking step, the point is our new head node, here disconnected, the output of this node on it (as already chain, thus previously been connected to the end of the list a).
code show as below:
class Solution(object):
def rotateRight(self, head, k):
if (not head) | (k == 0):
return head
elif not head.next:
return head
cur = head
length = 1
while cur.next:
cur = cur.next
length = length + 1
cur.next = head
k = k % length
if k == 0:
cur.next = None
return head
cur = head
for i in range(1, length - k):
cur = cur.next
head = cur.next
cur.next = None
return head
operation result
Execution Time: 24 ms
Memory consumption: 11.9 MB
