如何做网站插件,网站推广主要方法,怎么建设咨询网站,长沙百度关键词搜索记录了初步解题思路 以及本地实现代码#xff1b;并不一定为最优 也希望大家能一起探讨 一起进步 目录 8/14 617. 合并二叉树8/15 833. 字符串中的查找与替换8/16 2682. 找出转圈游戏输家8/17 1444. 切披萨的方案数8/18 1388. 3n 块披萨8/19 2235. 两整数相加8/20 8/14 617. 合…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 目录 8/14 617. 合并二叉树8/15 833. 字符串中的查找与替换8/16 2682. 找出转圈游戏输家8/17 1444. 切披萨的方案数8/18 1388. 3n 块披萨8/19 2235. 两整数相加8/20 8/14 617. 合并二叉树 dfs深搜 class TreeNode(object):def __init__(self, val0, leftNone, rightNone):self.val valself.left leftself.right right
def mergeTrees(root1, root2)::type root1: TreeNode:type root2: TreeNode:rtype: TreeNodedef func(n1,n2):if not n1:return n2if not n2:return n1node TreeNode(n1.valn2.val)node.left func(n1.left,n2.left)node.right func(n1.right,n2.right)return nodereturn 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: strfrom collections import defaultdictn len(s)op defaultdict(list)for i,ind in enumerate(indices):op[ind].append(i)ans []i 0while in:tag Falseif i in op:for ind in op[i]:if s[i:ilen(sources[ind])]sources[ind]:tag Trueans.append(targets[ind])ilen(sources[ind])breakif not tag:ans.append(s[i])i1return .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]*ncur 0i1while not do[cur]:do[cur]Truecuri*kcur%ni1return [i1 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: intmod 10**97m,nlen(pizza),len(pizza[0])apples [[0]*(n1) for _ in range(m1)]dp [[[0 for j in range(n)] for i in range(m)] for _ in range(k1)]for i in range(m-1,-1,-1):for j in range(n-1,-1,-1):apples[i][j] apples[i][j1]apples[i1][j]-apples[i1][j1](pizza[i][j]A)if apples[i][j]0:dp[1][i][j] 1 else:dp[1][i][j] 0for t in range(1,k1):for i in range(m):for j in range(n):for ii in range(i1,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(j1,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] 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: intdef func(slices):m len(slices)n (len(slices)1)//3dp [[float(-inf) for _ in range(n1)] 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,n1):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 def sum(num1, num2)::type num1: int:type num2: int:rtype: intreturn num1num2 8/20
