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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4/1 2810. 故障键盘

依次考虑 将字符串放入数组中
记录插入的位置如果逆序 则从头开始插入
loc记录是顺序还是逆序

def finalString(s):""":type s: str:rtype: str"""loc =Trueans=[]for c in s:if c=="i":loc = not locelif loc:ans.append(c)else:ans = [c]+ansif loc:return ''.join(ans)else:return ''.join(ans[::-1])

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4/2 894. 所有可能的真二叉树

真二叉树总结点必定为奇数
mem[i]记录i个节点的真二叉树情况

class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
def allPossibleFBT(n):""":type n: int:rtype: List[TreeNode]"""if n<3 or n%2==0:return []mem = {}mem[1] = [TreeNode(0)]def check(num):if num in mem:return mem[num]tmp = []for l in range(1,num-1,2):left = check(l)right = check(num-1-l)for ln in left:node = TreeNode(0)node.left = lnfor rn in right:node.right = rntmp.append(node)mem[num]=tmpreturn tmpreturn check(n)

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4/3 1379. 找出克隆二叉树中的相同节点

dfs搜索

class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
def getTargetCopy(original, cloned, target):""":type original: TreeNode:type cloned: TreeNode:type target: TreeNode:rtype: TreeNode"""def check(ori,clo):if not ori:return oriif ori==target:return cloleft = check(ori.left,clo.left)right = check(ori.right,clo.right)if left:return leftreturn rightreturn check(original, cloned)

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4/4 2192. 有向无环图中一个节点的所有祖先

如果x为y的祖先 那么x的祖先必定是y的祖先
cur存储祖先节点已经都被考虑过的节点
par[x]记录x当前未被考虑的祖先个数
ans[x]记录x的祖先

def getAncestors(n, edges):""":type n: int:type edges: List[List[int]]:rtype: List[List[int]]"""cur = set(list(range(n)))ans = [set() for _ in range(n)]m = [[] for _ in range(n)]par = [0]*nfor x,y in edges:if y in cur:cur.remove(y)m[x].append(y)par[y]+=1while cur:tmp = set()for p in cur:for child in m[p]:ans[child].add(p)ans[child]|=ans[p]par[child]-=1if par[child]==0:tmp.add(child)cur = tmpfor i in range(n):ans[i] = sorted(list(ans[i]))return ans

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4/5 1026. 节点与其祖先之间的最大差值

从根节点往子节点判断 记录当前最大值和最小值
当前节点与最大值最小值能够得到的最大差值

class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
def maxAncestorDiff(root):""":type root: TreeNode:rtype: int"""global ansans = 0def check(node,minv,maxv):global ansif not node:returnans = max(ans,abs(minv-node.val))ans = max(ans,abs(maxv-node.val))minv = min(node.val,minv)maxv = max(node.val,maxv)check(node.left,minv,maxv)check(node.right,minv,maxv)check(root,root.val,root.val)return ans

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4/6 1483. 树节点的第 K 个祖先

dp[i]用来记录i的祖先
因为n很大 使用二维dp[i][j] 记录i的第2^j个祖先
dp[i][0]即为i的父节点
dp[i][j] = dp[dp[i][j-1]][j-1]
即i的第2^j个祖先 是i的第2^(j-1) 个祖先的第2^(j-1)个祖先

class TreeAncestor(object):def __init__(self, n, parent):""":type n: int:type parent: List[int]"""self.dp = [[] for _ in range(n)]for i in range(n):self.dp[i].append(parent[i])j=1while True:tag = Truefor i in range(n):v = -1if self.dp[i][j-1]!=-1:v = self.dp[self.dp[i][j-1]][j-1]self.dp[i].append(v)if v!=-1:tag = Falseif tag:breakj+=1      def getKthAncestor(self, node, k):""":type node: int:type k: int:rtype: int"""ans = node pos = 0while k and ans!=-1:if pos>=len(self.dp[ans]):return -1if k&1:ans = self.dp[ans][pos]k = k>>1pos+=1return ans

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4/7 1600. 王位继承顺序

可以看做是一个多叉树
继承顺序为父-子 前序遍历 根左右
dead记录死亡的人
m[x]记录x的儿子

class ThroneInheritance(object):def __init__(self, kingName):""":type kingName: str"""from collections import defaultdictself.dead = set()self.king = kingNameself.m = defaultdict(list)def birth(self, parentName, childName):""":type parentName: str:type childName: str:rtype: None"""self.m[parentName].append(childName)def death(self, name):""":type name: str:rtype: None"""self.dead.add(name)def getInheritanceOrder(self):""":rtype: List[str]"""ans = []def pre(name):if name not in self.dead:ans.append(name)if name in self.m:for c in self.m[name]:pre(c)pre(self.king)return ans

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