家政服务技术支持东莞网站建设网络营销方法

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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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 = valself.left = leftself.right = right
def mergeTrees(root1, root2):""":type root1: TreeNode:type root2: TreeNode:rtype: TreeNode"""def func(n1,n2):if not n1:return n2if not n2:return n1node = TreeNode(n1.val+n2.val)node.left = func(n1.left,n2.left)node.right = func(n1.right,n2.right)return nodereturn func(root1,root2)

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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 defaultdictn = len(s)op = defaultdict(list)for i,ind in enumerate(indices):op[ind].append(i)ans = []i = 0while i<n:tag = Falseif i in op:for ind in op[i]:if s[i:i+len(sources[ind])]==sources[ind]:tag = Trueans.append(targets[ind])i+=len(sources[ind])breakif not tag:ans.append(s[i])i+=1return "".join(ans)

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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]*ncur = 0i=1while not do[cur]:do[cur]=Truecur+=i*kcur%=ni+=1return [i+1 for i in range(n) if not do[i]]

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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+7m,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] = 0for 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])%modfor 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])%modreturn dp[k][0][0]

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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)//3dp = [[float("-inf") for _ in range(n+1)] for _ in range(m)]dp[0][0] = 0dp[0][1] = slices[0]dp[1][0] = 0dp[1][1] = max(slices[0],slices[1])for i in range(2,m):dp[i][0] = 0for 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]))

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8/19 2235. 两整数相加

如题相加

def sum(num1, num2):""":type num1: int:type num2: int:rtype: int"""return num1+num2

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8/20

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