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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12/11 1631. 最小体力消耗路径

dijstra
最小堆 找到当前最小的路径

import heapq
def minimumEffortPath(heights):""":type heights: List[List[int]]:rtype: int"""m,n = len(heights),len(heights[0])dist = [0]+[float("inf")]*(m*n-1)q = [(0,0,0)]path =set()while q:d,x,y = heapq.heappop(q)idx = x*n+yif idx in path:continueif (x,y)==(m-1,n-1):breakpath.add(idx)for xx,yy in [(x-1,y),(x+1,y),(x,y-1),(x,y+1)]:if 0<=xx<m and 0<=yy<n:maxd = max(d,abs(heights[x][y]-heights[xx][yy]))if maxd<=dist[xx*n+yy]:dist[xx*n+yy] = maxdheapq.heappush(q,(dist[xx*n+yy],xx,yy))return dist[m*n-1]

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12/12 2454. 下一个更大元素 IV

将数值从大到小排序 并记录其坐标(num,i)
使用一个有序队列s记录已经处理过的坐标
对于当前处理的nums[x] s中所有数值都比其大 只要找到比x大的下下个元素值
即为nums[x]对应的第二大元素坐标

def secondGreaterElement(nums):""":type nums: List[int]:rtype: List[int]"""from sortedcontainers import SortedListl = [(x,i) for i,x in enumerate(nums)]l.sort(key=lambda x:-x[0])s = SortedList()n = len(nums)ans = [-1]*nfor _,i in l:loc = s.bisect_right(i)if loc+1<len(s):ans[i] = nums[s[loc+1]]s.add(i)return ans

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12/13 2697. 字典序最小回文串

左右指针依次比较 对应的位置需要相同且使用字典序小的字母

def makeSmallestPalindrome(s):""":type s: str:rtype: str"""l,r=0,len(s)-1li = [""]*len(s)while l<=r:c = s[l]if s[r]<s[l]:c=s[r]li[l]=li[r]=cl+=1r-=1return ''.join(li)

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12/14 2132. 用邮票贴满网格图

二维前缀和判断是否可以贴邮票
二维差分判断是否所有空格子被覆盖

def possibleToStamp(grid, stampHeight, stampWidth):""":type grid: List[List[int]]:type stampHeight: int:type stampWidth: int:rtype: bool"""m,n=len(grid),len(grid[0])psum = [[0]*(n+2) for _ in range(m+2)]diff = [[0]*(n+2) for _ in range(m+2)]for i in range(1,m+1):for j in range(1,n+1):psum[i][j] = psum[i-1][j]+psum[i][j-1]-psum[i-1][j-1]+grid[i-1][j-1]for i in range(1,m+2-stampHeight):for j in range(1,n+2-stampWidth):x = i+stampHeight-1y = j+stampWidth-1if psum[x][y]-psum[x][j-1]-psum[i-1][y]+psum[i-1][j-1]==0:diff[i][j]+=1diff[i][y+1]-=1diff[x+1][j]-=1diff[x+1][y+1]+=1for i in range(1,m+1):for j in range(1,n+1):diff[i][j] += diff[i-1][j]+diff[i][j-1]-diff[i-1][j-1]if diff[i][j]==0 and grid[i-1][j-1]==0:return Falsereturn True

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12/15 2415. 反转二叉树的奇数层

BFS

class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = rightdef reverseOddLevels(root):""":type root: Optional[TreeNode]:rtype: Optional[TreeNode]"""li = [root]lev = 0while len(li)>0:if lev%2==1:l,r=0,len(li)-1while l<r:x,y = li[l],li[r]x.val,y.val=y.val,x.vall+=1r-=1tmp=[]for node in li:if not node.left:breaktmp.append(node.left)tmp.append(node.right)lev+=1li = tmpreturn root

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12/16 2276. 统计区间中的整数数目

线段树
l,r为左右端点 cnt为区间内整数个数
如果当前区间全覆盖 则cnt=r-l+1

class CountIntervals(object):def __init__(self,l=1,r=10**9):self.left = self.right = Noneself.l,self.r,self.cnt=l,r,0def add(self, left, right):""":type left: int:type right: int:rtype: None"""if self.cnt == self.r-self.l+1:returnif left<=self.l and self.r<=right:self.cnt = self.r-self.l+1returnmid = (self.l+self.r)//2if self.left is None:self.left = CountIntervals(self.l,mid)if self.right is None:self.right = CountIntervals(mid+1,self.r)if left<=mid:self.left.add(left,right)if mid<right:self.right.add(left,right)self.cnt = self.left.cnt+self.right.cntdef count(self):""":rtype: int"""return self.cnt

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12/17 746. 使用最小花费爬楼梯

当前i楼梯可以由 i-1,i-2到达

def minCostClimbingStairs(cost):""":type cost: List[int]:rtype: int"""f1,f2=0,0for i in cost:f1,f2=i+min(f1,f2),f1ret = min(f1,f2)return ret

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