算法思路 linux 2023-01-19 Source Edit History undefined 二分查找 目标函数单调性(单调递增或者递减) 存在上下界(bounded) 能够通过索引访问(index accessible) 代码模版 12345678910left, right = 0, len(array)-1while left <= right: mid = (left + right) / 2; # 处理越界问题 int mid = left + (right-left)/2; if array[mid] == target: # find the target break or return result elif array[mid] < target: left = mid + 1; else: right = mid - 1; 了解牛顿迭代法