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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11/4 633. 平方数之和

1. 正常
   计算最大可能的平方数 依次减小判断
2. 双指针
   假设a从0开始,b从最大可能数开始 如果aa+bb
   小于c 那么增加a
   大于c 那么减小b

def judgeSquareSum(c):""":type c: int:rtype: bool"""import mathn = int(math.sqrt(c))for i in range(n,-1,-1):tmp = c-i*im = int(math.sqrt(tmp))if m*m==tmp:return Truereturn Falsedef judgeSquareSum2(c):""":type c: int:rtype: bool"""import mathr = int(math.sqrt(c))l = 0while l<=r:s = r*r+l*lif s==c:return Trueelif s<c:l+=1else:r-=1return False

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11/5 3222. 求出硬币游戏的赢家

115 = 75+4*10
取一次需要1个75 4个10
计算最多能取几次 如果是奇数次就是A赢 偶数次B赢

def losingPlayer(x, y):""":type x: int:type y: int:rtype: str"""return "Alice" if min(x,y//4)%2 else "Bob"

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11/6 3254. 长度为 K 的子数组的能量值 I

如果满足条件那么相邻的上升数值相差1
cnt统计以当前i为结尾有多少个连续上升的个数
如果个数大于等于k个那么以当前i为最后一个数可以得到满足条件的子数组

def resultsArray(nums, k):""":type nums: List[int]:type k: int:rtype: List[int]"""n=len(nums)cnt=0ans = [-1]*(n-k+1)for i in range(n):if i==0 or nums[i]-nums[i-1]!=1:cnt=1else:cnt+=1if cnt>=k:ans[i-k+1]=nums[i]return ans

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11/7 3255. 长度为 K 的子数组的能量值 II

如果满足条件那么相邻的上升数值相差1
cnt统计以当前i为结尾有多少个连续上升的个数
如果个数大于等于k个那么以当前i为最后一个数可以得到满足条件的子数组

def resultsArray(nums, k):""":type nums: List[int]:type k: int:rtype: List[int]"""n=len(nums)cnt=0ans = [-1]*(n-k+1)for i in range(n):if i==0 or nums[i]-nums[i-1]!=1:cnt=1else:cnt+=1if cnt>=k:ans[i-k+1]=nums[i]return ans

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11/8 3235. 判断矩形的两个角落是否可达

如果起点或终点在园内 不存在路径
如果园和左、上边相交 并且和右、下边相交 不存在路径

def canReachCorner(xCorner, yCorner, circles):""":type xCorner: int:type yCorner: int:type circles: List[List[int]]:rtype: bool"""def incircle(px,py,x,y,r):return (x-px)**2+(y-py)**2<=r**2def topleft(x,y,r,xc,yc):return (abs(x)<=r and 0<=y<=yc)or(0<=x<=xc and abs(y-yc)<=r)or(incircle(x, y, 0, yc, r))def bottomright(x,y,r,xc,yc):return (abs(y)<=r and 0<=x<=xc)or(0<=y<=yc and abs(x-xc)<=r)or(incircle(x, y, xc, 0, r))def circleincircle(x1,y1,r1,x2,y2,r2,xc,yc):return (x1 - x2) ** 2 + (y1 - y2) ** 2 <= (r1 + r2) ** 2 and x1 * r2 + x2 * r1 < (r1 + r2) * xCorner and y1 * r2 + y2 * r1 < (r1 + r2) * yCornern=len(circles)visited=[False]*ndef dfs(i):x1,y1,r1=circles[i]if bottomright(x1, y1, r1, xCorner, yCorner):return Truevisited[i]=Truefor j,(x2,y2,r2) in enumerate(circles):if (not visited[j]) and circleincircle(x1, y1, r1, x2, y2, r2, xCorner, yCorner) and dfs(j):return Truereturn Falsefor i,(x,y,r) in enumerate(circles):if incircle(0, 0, x, y, r) or incircle(xCorner, yCorner, x, y, r):return Falseif (not visited[i]) and topleft(x, y, r, xCorner, yCorner) and dfs(i):return Falsereturn True

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11/9 3242. 设计相邻元素求和服务

记录每个value的位置

class NeighborSum(object):def __init__(self, grid):""":type grid: List[List[int]]"""self.n=len(grid)self.grid=gridself.m = {}for i in range(self.n):for j in range(self.n):self.m[grid[i][j]]=(i,j)def adjacentSum(self, value):""":type value: int:rtype: int"""i,j = self.m[value]ans = 0if i-1>=0:ans+=self.grid[i-1][j]if i+1<self.n:ans+=self.grid[i+1][j]if j-1>=0:ans+=self.grid[i][j-1]if j+1<self.n:ans+=self.grid[i][j+1]return ansdef diagonalSum(self, value):""":type value: int:rtype: int"""i,j = self.m[value]ans = 0if i-1>=0 and j-1>=0:ans+=self.grid[i-1][j-1]if i+1<self.n and j+1<self.n:ans+=self.grid[i+1][j+1]if i+1<self.n and j-1>=0:ans+=self.grid[i+1][j-1]if i-1>=0 and j+1<self.n:ans+=self.grid[i-1][j+1]return ans

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11/10 540. 有序数组中的单一元素

在单一元素前 元素成对出现
偶数位和偶数位+1元素相同 单一元素位置为偶数
单一元素后 元素成对出现 偶数位和偶数位-1元素相同
二分查找偶数位元素

def singleNonDuplicate(nums):""":type nums: List[int]:rtype: int"""l,r = 0,len(nums)-1while l<r:mid = (l+r)//2if mid%2==1:mid-=1if nums[mid]==nums[mid+1]:l = mid+2else:r = midreturn nums[l]

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