leetcode-python 第十二周

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最近在看course的题目，都是些简单的题目，每天花在上面的时间不多。

1.Delete Node In a Linked List [52ms]

# 方法1：双指针遍历，用后面的值代替前面的值class Solution(object):    def deleteNode(self, node):        """        :type node: ListNode        :rtype: void Do not return anything, modify node in-place instead.        """        p = node        q = node.next        while q != None:            p.val = q.val            if q.next == None:                p.next = None                del q                break            p = p.next            q = q.next# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None

2. Insert Of Two Array II [39ms]

# 方法1：二指针class Solution(object):    def intersect(self, nums1, nums2):        """        :type nums1: List[int]        :type nums2: List[int]        :rtype: List[int]        """        low1, high1 = 0, len(nums1)        low2, high2 = 0, len(nums2)        nums1.sort()        nums2.sort()        ans = []        while low1 < high1 and low2 < high2:            if nums1[low1] == nums2[low2]:                ans.append(nums1[low1])                low1 += 1                low2 += 1            elif nums1[low1] > nums2[low2]:                low2 += 1            else:                low1 += 1        return ans

3. Merge Two Sorted Lists [72ms]

class Solution(object):    def mergeTwoLists(self, l1, l2):        """        :type l1: ListNode        :type l2: ListNode        :rtype: ListNode        """        if l1 == None: return l2        if l2 == None: return l1        if l1 != None and l2 != None:            if l1.val > l2.val:                tmp = ListNode(l2.val)                tmp.next = l1                l1 = tmp                l2 = l2.next        p1, p2 = l1, l1.next        while p2 != None and l2 != None:            if p2.val < l2.val:                p1, p2 = p1.next, p2.next            else:                tmp = ListNode(l2.val)                p1.next = tmp                p1.next.next = p2                p1 = p1.next                l2 = l2.next        if l2 != None:            p1.next = l2        return l1# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None

4. Swqp Nodes in Pairs [39ms]

class Solution(object):    def swapPairs(self, head):        """        :type head: ListNode        :rtype: ListNode        """        if head is None or head.next is None:            return head        p, q = head, head.next        p.next = q.next        q.next = p        head = q        p, q = q, p        while q.next != None and q.next.next != None:            p = q.next            q.next = p.next            p.next = p.next.next            q.next.next = p            p, q = q, p        return head# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None

5. Reversed Linked List [52ms]

# 方法1：开辟空间先储存在新建class Solution(object):    def reverseList(self, head):        """        :type head: ListNode        :rtype: ListNode        """        if head == None:            return head        val = []        p = head        while p != None:            val.append(p.val)            p = p.next        rhead = ListNode(val[-1])        q = rhead        for i in range(-2, -len(val)-1, -1):            tmp = ListNode(val[i])            q.next = tmp            q = q.next        return rhead# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None

6. Maximum Depth of Binary Tree [146ms]

class Solution(object):    def maxDepth(self, root):        """        :type root: TreeNode        :rtype: int        """        if root == None:            return 0        ans = self.goDown(root, 0)        return ans    def goDown(self, node, n):        if node == None:            return n        else:            return max(self.goDown(node.left, n+1), self.goDown(node.right, n+1))# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = None

7. Linked List Cycle [82ms]

# 方法1：快慢指针，刚开始以为它只会检测一定长度的环，# 但是由于单向链表的结构，所以循环多几次就可以找到，毕竟快指针永远快一格# 注意为空条件class Solution(object):    def hasCycle(self, head):        """        :type head: ListNode        :rtype: bool        """        slow = head        fast = head        while slow != None and fast != None:            slow = slow.next            if fast.next != None:                fast = fast.next.next            else:                return False            if slow == fast:                return True        return False# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = None

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