题解 | 旋转数组的最小数字
旋转数组的最小数字
https://www.nowcoder.com/practice/9f3231a991af4f55b95579b44b7a01ba
# # 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可 # # # @param nums int整型一维数组 # @return int整型 # class Solution: def minNumberInRotateArray(self , nums: List[int]) -> int: # write code here if len(nums) == 0: return 0 i, j = 0, len(nums)-1 while i < j: m = (i + j ) // 2 if nums[m] > nums[j]: i = m + 1 elif nums[m] < nums[j]: j = m else: j -= 1 return nums[i]