【经典LeetCode算法题目专栏分类】【第6期】二分查找系列:x的平方根、有效完全平方数、搜索二位矩阵、寻找旋转排序数组最小值

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X 的平方根

|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def mySqrt****(**** self****,**** x****:**** int ) -> int : l****,**** r****,**** ans = 0****,**** x****,**** - 1 while l <= r****:**** mid = ( l + r****)**** // 2 if mid * mid <= x****:**** ans = mid l = mid + 1 else : r = mid - 1 return ans |

有效完全平方数

|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def isPerfectSquare****(**** self****,**** num****:**** int ) -> bool : l = 0 r = num while l <= r****:**** mid = ( l****+**** r****)//**** 2 if mid * mid == num****:**** return True elif mid * mid < num****:**** l = mid + 1 else : r = mid - 1 return False |

搜索旋转排序数组

|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def search****(**** self****,**** nums****:**** List****[**** int ], target****:**** int ) -> int : if not nums****:**** return - 1 # 二分法 n = len ( nums****)**** left = 0 right = n - 1 while left <= right****:**** mid = ( left + right****)**** // 2 if nums****[**** mid****]**** == target****:**** return mid if nums****[**** 0****]**** <= nums****[**** mid****]:**** # 说明左边有序 if nums****[**** 0****]**** <= target < nums****[**** mid****]:**** right = mid - 1 else : left = mid + 1 else : # 右边有序 if nums****[**** mid****]**** < target <= nums****[**** n****-**** 1****]:**** left = mid + 1 else : right = mid - 1 return - 1 |

搜索二位矩阵

|------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def searchMatrix****(**** self****,**** matrix****:**** List****[**** List****[**** int ]], target****:**** int ) -> bool : if not matrix****:**** return False # 二分查找 row = index // n ; col = index % n m = len ( matrix****)**** n = len ( matrix****[**** 0****])**** left = 0 right = m * n - 1 while left <= right****:**** mid = ( left + right****)**** // 2 cur_m = mid // n cur_n = mid % n if matrix****[**** cur_m****][**** cur_n****]**** == target****:**** return True elif matrix****[**** cur_m****][**** cur_n****]**** > target****:**** right = mid - 1 else : left = mid + 1 return False |

搜索二维矩阵2

|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| def searchMatrix****(**** self****,**** matrix****,**** target****):**** """ :type matrix: List[List[int]] :type target: int :rtype: bool """ # 1.暴力法 for i in range(m) for j in range(n) O(mn) # 2.剪枝搜索,假设从左下角开始搜索O(m+n) if not matrix****:**** return False m = len ( matrix****)**** n = len ( matrix****[**** 0****])**** row = m - 1 col = 0 while row >= 0 and col < n****:**** if matrix****[**** row****][**** col****]**** > target****:**** row -= 1 elif matrix****[**** row****][**** col****]**** < target****:**** col += 1 else : return True return False |

寻找旋转排序数组中的最小值

|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|
| class Solution : def findMin****(**** self****,**** nums****:**** List****[**** int ]) -> int : if len ( nums****)**** == 1****:**** return nums****[**** 0****]**** left = 0 right = len ( nums****)**** - 1 while left < right****:**** mid = ( left + right****)**** // 2 if nums****[**** mid****]**** < nums****[**** right****]:**** # mid可能是最小值 right = mid else : # mid一定不是最小值 left = mid + 1 return nums****[**** left****]**** |

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