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

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3/11 2129. 将标题首字母大写

从头按规则依次判断
curLen记录当前字母个数
如果遇到空格时少于等于两个则特殊处理
start记录下一个字母是否是首字母

def capitalizeTitle(title):""":type title: str:rtype: str"""start = TruecurLen = 0ans = []for c in title:if c!=" ":if start:ans.append(c.upper())else:ans.append(c.lower())curLen+=1start = Falseelse:start = Trueif curLen<=2:for i in range(curLen):ans[-1-i] = ans[-1-i].lower()curLen=0ans.append(" ")if curLen<=2:for i in range(curLen):ans[-1-i] = ans[-1-i].lower()return ''.join(ans)

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3/12 1261. 在受污染的二叉树中查找元素

根据树结构还原
并且存储出现过的数值

class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightclass FindElements(object):def __init__(self, root):""":type root: TreeNode"""self.m = {}root.val = 0self.root = rootl = [root]while l:tmp = []for node in l:v = node.valself.m[v]=1if node.left:node.left.val = v*2+1tmp.append(node.left)if node.right:node.right.val = v*2+2tmp.append(node.right)l=tmp[:]def find(self, target):""":type target: int:rtype: bool"""return target in self.m

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3/13 2864. 最大二进制奇数

二进制奇数说明最低位必定为1
统计字符串中1的个数 最低位为1 剩余从高到低往下排

def maximumOddBinaryNumber(s):""":type s: str:rtype: str"""n=len(s)m = s.count("1")return  "1"*(m-1)+"0"*(n-m)+"1"

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3/14 2789. 合并后数组中的最大元素

从后往前将元素依次放入一个单调递增的栈
如果当前元素小于等于栈顶元素 说明可以合并

def maxArrayValue(nums):""":type nums: List[int]:rtype: int"""l = []for num in nums[::-1]:if not l or l[-1]<num:l.append(num)else:l[-1]+=numreturn max(l)

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3/15 2312. 卖木头块

dp
使用dp[i][j]来存储i*j能够得到的值
按竖着横着切开

def sellingWood(m, n, prices):""":type m: int:type n: int:type prices: List[List[int]]:rtype: int"""dp = [[0]*(n+1) for _ in range(m+1)]for x,y,v in prices:dp[x][y]=vfor i in range(1,m+1):for j in range(1,n+1):for k in range(1,j//2+1):dp[i][j] = max(dp[i][j],dp[i][k]+dp[i][j-k])for k in range(1,i//2+1):dp[i][j] = max(dp[i][j],dp[k][j]+dp[i-k][j])return dp[m][n]

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3/16 2684. 矩阵中移动的最大次数

每一步必定往右挪动
dfs 从第一列往后走

def maxMoves(grid):""":type grid: List[List[int]]:rtype: int"""n,m=len(grid),len(grid[0])mem = {}def dfs(i,j):if (i,j) in mem:return mem[(i,j)]v = 0if j+1<m:if grid[i][j+1]>grid[i][j]:v = max(v,1+dfs(i,j+1))if i-1>=0 and grid[i-1][j+1]>grid[i][j]:v = max(v,1+dfs(i-1,j+1))if i+1<n and grid[i+1][j+1]>grid[i][j]:v = max(v,1+dfs(i+1,j+1))mem[(i,j)] = vreturn vans = 0for i in range(n):ans = max(ans,dfs(i,0))return ans

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3/17

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