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 rotateArray int整型一维数组 # @return int整型 # class Solution: def minNumberInRotateArray(self , rotateArray: List[int]) -> int: left = 0 right = len(rotateArray)-1 while left < right: mid = (left + right) // 2 if rotateArray[mid] == rotateArray[right]: # 当中间值等于右界值时,删除右界值 right -= 1 elif rotateArray[mid] > rotateArray[right]: # 等号不能并在这里,当数组为[10111]时会出错 left = mid + 1 # 当中间值小于rotateArray[mid-1],且小于右界值时,rotateArray[mid]为最小,直接输出 elif rotateArray[mid] < rotateArray[right] and rotateArray[mid] < rotateArray[mid-1]: return rotateArray[mid] else: # 当中间值小于右界值,大于前一个值时 right = mid-1 return rotateArray[left]