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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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3/31 2278. 字母在字符串中的百分比

遍历 求出字母个数

def percentageLetter(s, letter):""":type s: str:type letter: str:rtype: int"""num = 0n=len(s)for c in s:if c==letter:num+=1return num*100//n

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4/1 2140. 解决智力问题

一眼动归
因为对于某个位置i是否可以解决取决于他前面的问题 所以从后往前遍历更加方便
定义dp[i]为解决i~n之间的问题可以得到的最高分数
dp[i]=points[i]+dp[i+brainpower[i]+1]

def mostPoints(questions):""":type questions: List[List[int]]:rtype: int"""n=len(questions)dp=[0]*(n+1)for i in range(n-1,-1,-1):dp[i]=max(dp[i+1],questions[i][0]+dp[min(n,i+questions[i][1]+1)])return dp[0]

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4/2 2873. 有序三元组中的最大值 I

1.遍历 三重循环
2.为了使值最大 对于每个j 要找到它左边的最大值nums[i] 和右边的最大值nums[k]
lmax[j],rmax[j]分别记录j左右的最大值

def maximumTripletValue(nums):""":type nums: List[int]:rtype: int"""cur = 0n=len(nums)for i in range(n-2):for j in range(i+1,n-1):for k in range(j+1,n):cur=max(cur,(nums[i]-nums[j])*nums[k])return curdef maximumTripletValue2(nums):""":type nums: List[int]:rtype: int"""n=len(nums)lmax=[0]*nrmax=[0]*nfor i in range(1,n):lmax[i]=max(lmax[i-1],nums[i-1])rmax[n-1-i]=max(rmax[n-i],nums[n-i])cur=0for j in range(1,n-1):cur=max(cur,(lmax[j]-nums[j])*rmax[j])return cur

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4/3 2874. 有序三元组中的最大值 II

数组长度很大不能循环遍历
为了使值最大 对于每个j 要找到它左边的最大值nums[i] 和右边的最大值nums[k]
lmax[j],rmax[j]分别记录j左右的最大值

def maximumTripletValue(nums):""":type nums: List[int]:rtype: int"""n=len(nums)lmax=[0]*nrmax=[0]*nfor i in range(1,n):lmax[i]=max(lmax[i-1],nums[i-1])rmax[n-1-i]=max(rmax[n-i],nums[n-i])cur=0for j in range(1,n-1):cur=max(cur,(lmax[j]-nums[j])*rmax[j])return cur

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4/4 1123. 最深叶节点的最近公共祖先

dfs

class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef lcaDeepestLeaves(root):""":type root: Optional[TreeNode]:rtype: Optional[TreeNode]"""global ans,maxdans=Nonemaxd = -1def dfs(node,dep):global ans,maxdif not node:maxd = max(maxd,dep)return depleft = dfs(node.left,dep+1)right = dfs(node.right,dep+1)if left==right==maxd:ans = nodereturn max(left,right)dfs(root,0)return ans

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4/5 1863. 找出所有子集的异或总和再求和

遍历

def subsetXORSum(nums):""":type nums: List[int]:rtype: int"""ans=0n=len(nums)for i in range(1<<n):tmp=0for j in range(n):if i&(1<<j):tmp^=nums[j]ans+=tmpreturn ans

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4/6 368. 最大整除子集

1. 遍历
   ans内每个num结尾都一个备选答案
   每一个数 遍历当前备选答案 保留长度最大的一个为当前Num的备选答案
   2.dp
   dp[i] nums[i]最大子集个数

def largestDivisibleSubset(nums):""":type nums: List[int]:rtype: List[int]"""n = len(nums)nums.sort()ans = [[nums[0]]]for i in range(1,n):c = nums[i]ret = []for j in range(len(ans)-1,-1,-1):tmp = ans[j][:]if c%tmp[-1]==0:tmp.append(c)if len(tmp)>len(ret):ret = tmp[:]if not ret:ret = [c]ans.append(ret)ans.sort(key=lambda x: len(x))return ans[-1]def largestDivisibleSubset2(nums):""":type nums: List[int]:rtype: List[int]"""n=len(nums)nums.sort()dp = [1]*nmsize,mval=1,1for i in range(n):for j in range(i):if nums[i]%nums[j]==0 and dp[j]+1>dp[i]:dp[i]=dp[j]+1if dp[i]>msize:msize=dp[i]mval = nums[i]if msize==1:return nums[:1]ret = []for i in range(n-1,-1,-1):if msize==0:return retif dp[i]==msize and mval%nums[i]==0:ret.append(nums[i])mval = nums[i]msize-=1return ret

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