给定一个已按照升序排列 的有序数组,找到两个数使得它们相加之和等于目标数。函数应该返回这两个下标值 index1 和 index2,其中 index1 必须小于 index2html
Two Sum II - Input array is sortedgit
# 解1,时间复杂度O(n),空间复杂度O(1) class Solution1: def twoSum(self, numbers: List[int], target: int) -> List[int]: l,r = 0,len(numbers)-1 while l < r: if numbers[l] + numbers[r] == target: return [l + 1, r + 1] elif numbers[l] + numbers[r] < target: l += 1 else: r -= 1 return [] # 解2,时间复杂度O(nlogn),空间复杂度O(1) class Solution2: def twoSum(self, numbers: List[int], target: int) -> List[int]: for n in range(len(numbers)): l,r = n + 1,len(numbers) - 1 tmp = target - numbers[n] while l <= r : mid = l + (r - l) // 2 if numbers[mid] == tmp: return [n + 1,mid + 1] elif numbers[mid] < tmp: l = mid + 1 else: r = mid - 1 return []
appium 启动时报错,An unknow server-side error occurred while processing the command...
换成UiAutomator1启动正常,怀疑是UiAutomator2问题。从新安装appium问题已解决(未能定位到确切缘由)
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