记录了初步解题思路 以及本地实现代码;并不一定为最优 也希望大家能一起探讨 一起进步
8/14 617. 合并二叉树
dfs深搜
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def mergeTrees(root1, root2):
"""
:type root1: TreeNode
:type root2: TreeNode
:rtype: TreeNode
"""
def func(n1,n2):
if not n1:
return n2
if not n2:
return n1
node = TreeNode(n1.val+n2.val)
node.left = func(n1.left,n2.left)
node.right = func(n1.right,n2.right)
return node
return func(root1,root2)
8/15 833. 字符串中的查找与替换
op存放该位置能替换的数值
从头遍历每个位置
def findReplaceString(s, indices, sources, targets):
"""
:type s: str
:type indices: List[int]
:type sources: List[str]
:type targets: List[str]
:rtype: str
"""
from collections import defaultdict
n = len(s)
op = defaultdict(list)
for i,ind in enumerate(indices):
op[ind].append(i)
ans = []
i = 0
while i<n:
tag = False
if i in op:
for ind in op[i]:
if s[i:i+len(sources[ind])]==sources[ind]:
tag = True
ans.append(targets[ind])
i+=len(sources[ind])
break
if not tag:
ans.append(s[i])
i+=1
return "".join(ans)
8/16 2682. 找出转圈游戏输家
模拟
def circularGameLosers(n, k):
"""
:type n: int
:type k: int
:rtype: List[int]
"""
do = [False]*n
cur = 0
i=1
while not do[cur]:
do[cur]=True
cur+=i*k
cur%=n
i+=1
return [i+1 for i in range(n) if not do[i]]
8/17 1444. 切披萨的方案数
动态规划 dp[k][i][j] 表示把坐标(i,j)右下方切割成k块的方案
def ways(pizza, k):
"""
:type pizza: List[str]
:type k: int
:rtype: int
"""
mod = 10**9+7
m,n=len(pizza),len(pizza[0])
apples = [[0]*(n+1) for _ in range(m+1)]
dp = [[[0 for j in range(n)] for i in range(m)] for _ in range(k+1)]
for i in range(m-1,-1,-1):
for j in range(n-1,-1,-1):
apples[i][j] = apples[i][j+1]+apples[i+1][j]-apples[i+1][j+1]+(pizza[i][j]=='A')
if apples[i][j]>0:
dp[1][i][j] = 1
else:
dp[1][i][j] = 0
for t in range(1,k+1):
for i in range(m):
for j in range(n):
for ii in range(i+1,m):
if apples[i][j]>apples[ii][j]:
dp[t][i][j] = (dp[t][i][j]+dp[t-1][ii][j])%mod
for jj in range(j+1,n):
if apples[i][j]>apples[i][jj]:
dp[t][i][j] = (dp[t][i][j]+dp[t-1][i][jj])%mod
return dp[k][0][0]
8/18 1388. 3n 块披萨
可转换为在3n个数中 选择n个不相邻的数 和最大
动态规划dp[i][j]表示前i个数选择j个不相邻的数 最大和
def maxSizeSlices(slices):
"""
:type slices: List[int]
:rtype: int
"""
def func(slices):
m = len(slices)
n = (len(slices)+1)//3
dp = [[float("-inf") for _ in range(n+1)] for _ in range(m)]
dp[0][0] = 0
dp[0][1] = slices[0]
dp[1][0] = 0
dp[1][1] = max(slices[0],slices[1])
for i in range(2,m):
dp[i][0] = 0
for j in range(1,n+1):
dp[i][j] = max(dp[i-1][j],dp[i-2][j-1]+slices[i])
return dp[m-1][n]
return max(func(slices[1:]),func(slices[0:-1]))
8/19 2235. 两整数相加
如题相加文章来源:https://www.toymoban.com/news/detail-658269.html
def sum(num1, num2):
"""
:type num1: int
:type num2: int
:rtype: int
"""
return num1+num2
8/20 2236. 判断根结点是否等于子结点之和
如题判断文章来源地址https://www.toymoban.com/news/detail-658269.html
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
def checkTree(root):
"""
:type root: Optional[TreeNode]
:rtype: bool
"""
return root.val==root.left.val+root.right.val
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