A33191. $下面 search 函数的平均时间复杂度为( )。int search(int n, int * p, int target) { int low = 0, high = n; while (low <= high) { int middle = (low + high) / 2; if (target == p[middle]) { return middle; } else if (targ\dots$

单选题困难

知识点

二分查找时间复杂度循环结构

题目描述

下面 search 函数的平均时间复杂度为( )。 int search(int n, int * p, int target) {int low = 0, high = n; while (low <= high) {int middle = (low + high) / 2; if (target == p[middle]) {return middle;} else if (target > p[middle]) {low = middle + 1;} else {high = middle - 1;}} return -1;}

O(n)

O(log(n))

O(1)

可能无法返回

上一题下一题