[leetcode]初级算法——链表

删除链表中的节点

Write a function to delete a node (except the tail) in a singly linked list, given only access to that node.

Example:

Input: head = [4,5,1,9], node = 5
Output: [4,1,9]
Explanation: You are given the second node with value 5, the linked list
             should become 4 -> 1 -> 9 after calling your function.

Code(By myself):

class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        node.val = node.next.val
        node.next = node.next.next
        return
        

Code(others):

class Solution:
    def deleteNode(self, node):
        """
        :type node: ListNode
        :rtype: void Do not return anything, modify node in-place instead.
        """
        if (not node) or (not node.next):
            return 
        node.val = node.next.val
        node.next = node.next.next

总结:

覆盖掉该节点

删除链表中的节点

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Code(By myself):

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

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        p = head
        q = head
        while n:
            q = q.next
            n -= 1
        if q == None:
            head = head.next
        while q != None and q.next != None:
            p = p.next
            q = q.next
        if p.next == None:
            head = None
        else:
            p.next = p.next.next
        return head

Code(others):

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: ListNode
        """
        if not head:
            return None
        first, second = head, head
        for i in range(n):
            if not first.next:
                return head.next
            first = first.next
        while first.next:
            first = first.next
            second = second.next
        second.next = second.next.next
        return head

总结:

学会优化代码

反转链表

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Code(By myself):

class Solution:
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        p = head
        if p == None or p.next == None:
            return head
        q = head.next
        if q.next == None:
            p.next = None
            q.next = p
            head = q
            return head
        o = q.next
        while o != None:
            q.next = p
            p = q
            q = o
            o = o.next
        q.next = p
        head.next = None
        head = q
        return head

Code(others):

class Solution:
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        
        new_head=None
        while head:
            p=head
            head=head.next
            p.next=new_head
            new_head=p
            
        return new_head

总结:

循环里的每一步都整理出来代码能更简洁

合并两个有序链表

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

Code(By myself):

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        p = ListNode(0)
        head = p
        while l1 and l2:
            if l1.val < l2.val:
                p.next = l1
                p = l1
                l1 = l1.next
            else:
                p.next = l2
                p = l2
                l2 =l2.next
        if l1:
            p.next = l1
        if l2:
            p.next = l2
        return head.next

回文链表

Given a singly linked list, determine if it is a palindrome.

Example:

Input: 1->2
Output: false

Code(By myself):

class Solution:
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        p = head
        list = []
        while p:
            list.append(p.val)
            p = p.next
        list.reverse()
        p = head
        for i in range(len(list)//2):
            if p.val != list[i]:
                return False
            p = p.next
        return True

环形链表

Given a linked list, determine if it has a cycle in it.

Code(By myself):

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        fast = head
        slow = head
        while fast and fast.next:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                return True
        return False
        

Code(others):

class Solution(object):
    def hasCycle(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        curr = head
        while curr is not None:
            nextnode = curr.next
            if nextnode is head:
                return True
            curr.next = head
            curr = nextnode
        return False

总结:

将节点next指向head形成一种标记

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.