leetcode刷题日记-热题100（4）
215-数组中第k个最大元素（python）
#快速选择算法-改进快排
class Solution:
def findKthLargest(self, nums: List[int], k: int) -> int:
def select(left,right,k_smallest):
if left==right: return nums[left]
pivot_index=random.randint(left,right)
pivot_index=partion(left,right,pivot_index)
if pivot_index==k_smallest:
return nums[k_smallest]
elif pivot_index>k_smallest:
return select(left,pivot_index-1,k_smallest)
else:
return select(pivot_index+1,right,k_smallest)
def partion(left,right,pivot_index):
pivot=nums[pivot_index]
nums[pivot_index],nums[right]=nums[right],nums[pivot_index]
ss_index=left
for i in range(left,right):
if nums[i]<pivot:
nums[ss_index],nums[i]=nums[i],nums[ss_index]
ss_index+=1
nums[right],nums[ss_index]=nums[ss_index],nums[right]
return ss_index
return select(0,len(nums)-1,len(nums)-k)
221-最大正方形（python）
#动态规划
class Solution:
def maximalSquare(self, matrix: List[List[str]]) -> int:
if not matrix:return 0
row=len(matrix)
col=len(matrix[0])
dp=[[0]*(col+1) for _ in range(row+1)]
res=0
for i in range(1,row+1):
for j in range(1,col+1):
if matrix[i-1][j-1]=='1':
dp[i][j]=min(dp[i-1][j-1],dp[i][j-1],dp[i-1][j])+1
res=max(res,dp[i][j]**2)
return res
236-二叉树的最近公共祖先（python）
#递归 分为均在右子树 均在左子树以及两边
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if not root:
return None
if root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left and right:
return root
return left if left else right
238-除自身外数组的乘积（python）
#左右积
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
res=[1]
p=q=1
for i in range(len(nums)-1):
p*=nums[i]
res.append(p)
for j in range(len(nums)-1,0,-1):
q*=nums[j]
res[j-1]*+q
return res
239-滑动窗口的最大值（python）
#滑动窗口
class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
if not nums:
return []
window, res = [], []
for i, num in enumerate(nums):
# 窗口滑动时的规律
if i>=k and window[0] <= i-k:
window.pop(0)
# 把最大值左边的数小的清除
while window and nums[window[-1]] <= num:
window.pop()
window.append(i)
if i >= k-1:
res.append(nums[window[0]])
return res
240-搜索二维矩阵（python）
class Solution:
def searchMatrix(self,matrix,target):
row=len(matrix)
col=len(matrix[0])
i,j=0,col-1
while i<row and j >=0:
if matrix[i][j]>target:
j-=1
elif matrix[i][j]<target:
i+=1
else:
return True
return False
279-完全平方数（python）
#四平方定理
class Solution:
def numSquares(self, n: int) -> int:
import math
while n % 4 == 0: n = n // 4
if n % 8 == 7: return 4
if int(math.sqrt(n)) ** 2 == n: return 1
i = 1
while i*i <= n:
j = math.sqrt(n - i*i)
if int(j) == j: return 2
i += 1
return 3
283-移动0至末尾（python）
#双指针
class Solution:
def moveZeroes(self, nums: List[int]) -> None:
j=0
for i in range(len(nums)):
if nums[i]!=0:
nums[i],nums[j]=nums[i],nums[j]
j+=1
return nums
287-寻找重复数（python）
#排序后比较
class Solution:
def findDuplicate(self,nums):
nums.sort()
for i in range(1,len(nums)):
if nums[i]==nums[i-1]:
return nums[i]
297-二叉树的序列化与反序列化（python）
class Codec:
def serialize(self, root):
def rserialize(root, string):
if root is None:
string += 'None,'
else:
string += str(root.val) + ','
string = rserialize(root.left, string)
string = rserialize(root.right, string)
return string
return rserialize(root, '')
def deserialize(self, data):
def rdeserialize(l):
if l[0] == 'None':
l.pop(0)
return None
root = TreeNode(l[0])
l.pop(0)
root.left = rdeserialize(l)
root.right = rdeserialize(l)
return root
data_list = data.split(',')
root = rdeserialize(data_list)
return root
300-最长上升子序列（python）
class Solution:
def lengthOfLIS(self, nums: List[int]) -> int:
if not nums:
return 0
dp=[1]*len(nums)
for i in range(len(nums)):
for j in range(i):
if nums[j]<nums[i]:
dp[i]=max(dp[i],dp[j]+1)
return max(dp)
322-零钱兑换（python）
class Solution(object):
def coinChange(self, coins, amount):
dp = [amount + 1 for _ in range(amount + 1)]
dp[0] = 0
# 到amount总金额(金额当索引)
for i in range(1, amount + 1):
for j in range(len(coins)):
# 只有硬币小于当前金额, 才能进行递推
if coins[j] <= i:
# dp[i]要么等于自身(已经被计算过),
# 要么等于coins[j](一次) + (i - coins[j])的次数
dp[i] = min(dp[i], dp[i-coins[j]] + 1)
return -1 if dp[amount] > amount else dp[amount]