随笔分类 - leetcode
摘要：class Solution: def maxSubArray(self, nums: List[int]) -> int: # 设当前的和为 cur_sum cur_sum = 0 res = nums[0] for num in nums: # 如果当前和大于0 if cur_sum > 0:
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def hasCycle(self, h
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摘要：class Solution: def maxArea(self, height: List[int]) -> int: # 双指针，每次移动短指针 i = 0 j = len(height) - 1 res = 0 while i <= j: if height[i] < height[j]: s
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摘要：class Solution: def majorityElement(self, nums: List[int]) -> int: # 摩尔计数法，1、互相抵消 2、计数阶段，不等于0则是重复最多的那个 count = 0 # 票数 res = None for n in nums: if cou
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摘要：# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Co
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摘要：# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution:
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摘要：# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution:
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摘要：# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution:
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摘要：# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution:
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摘要：# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution:
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摘要：# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution:
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摘要：""" # Definition for a Node. class Node: def __init__(self, val, left=None, right=None): self.val = val self.left = left self.right = right """ class
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摘要：def merge_sort(a, left, right): if left == right: return [a[left]] mid = left + (right-left) // 2 l1 = merge_sort(a, left, mid) l2 = merge_sort(a, mid
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def
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摘要：class Solution: def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode: cur = dummy = ListNode(0) while l1 and l2: if l1.val < l2.val: cur.ne
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def removeDuplicateN
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def deleteNode(self,
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def getKthFromEnd(se
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def getIntersectionN
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