二分查找[64-69]
时间复杂度O(log n)，要想到二分排序
35.搜索插入位置
class Solution:def searchInsert(self, nums: List[int], target: int) -> int:left = 0right = len(nums)-1while left <= right: #左闭右闭mid = (left+right)//2if nums[mid] < target:left = mid + 1elif nums[mid] > target:right = mid -1else: return midreturn mid+1 if nums[mid] < target else mid
74.搜索二维矩阵
class Solution:def searchMatrix(self, matrix: List[List[int]], target: int) -> bool:m = len(matrix) #行n = len(matrix[0]) #列left = 0right = m*n -1while left <= right:mid = (left+right)//2i = mid // nj = mid % nif matrix[i][j] < target:left = mid + 1elif matrix[i][j] > target:right = mid - 1else:return Truereturn False
34.在排序数组中查找元素的第一个和最后一个位置
class Solution:def searchRange(self, nums: List[int], target: int) -> List[int]:if len(nums) == 0: return [-1,-1]left = 0right = len(nums)-1while left <= right:mid = (left+right)//2if nums[mid] < target:left = mid +1elif nums[mid] > target:right = mid-1else :i = j = midwhile i>=0 and nums[i] == target:i -= 1while j<len(nums) and nums[j] == target:j += 1return [i+1, j-1]return [-1,-1]
# 1、首先，在 nums 数组中二分查找得到第一个大于等于 target的下标leftBorder；
# 2、在 nums 数组中二分查找得到第一个大于等于 target+1的下标， 减1则得到rightBorder；
# 3、如果开始位置在数组的右边或者不存在target，则返回[-1, -1] 。否则返回[leftBorder, rightBorder]
class Solution:def searchRange(self, nums: List[int], target: int) -> List[int]:def binarySearch(nums:List[int], target:int) -> int:left, right = 0, len(nums)-1while left<=right: # 不变量：左闭右闭区间middle = left + (right-left) //2 if nums[middle] >= target: right = middle - 1else: left = middle + 1return left # 若存在target，则返回第一个等于target的值 leftBorder = binarySearch(nums, target) # 搜索左边界rightBorder = binarySearch(nums, target+1) -1 # 搜索右边界if leftBorder == len(nums) or nums[leftBorder]!= target: # 情况一和情况二return [-1, -1]return [leftBorder, rightBorder]
33.搜索旋转排序数组
class Solution:def search(self, nums: List[int], target: int) -> int:left = 0right = len(nums)-1while left<= right:mid = (left+right)//2 #mid要么左边升序，要么右边升序if nums[mid] == target: return midelif nums[mid] < nums[right]: #[mid,right]右边升序if target > nums[mid] and target <= nums[right]: #target在区间（mid,right]left = mid +1else: right = mid-1else: #[left, mid],左边升序if target >= nums[left] and target < nums[mid]: #target在区间[left, mid）right = mid -1else: left = mid +1 return -1
153.寻找旋转排序数组中的最小值
