search-insert-position

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Algorithms

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You must&nbsp;write an algorithm with&nbsp;<code>O(log n)</code> runtime complexity.

&nbsp;
Example 1:

<pre>
Input: nums = [1,3,5,6], target = 5
Output: 2
</pre>

Example 2:

<pre>
Input: nums = [1,3,5,6], target = 2
Output: 1
</pre>

Example 3:

<pre>
Input: nums = [1,3,5,6], target = 7
Output: 4
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 104</code></li>
	<li><code>-104 &lt;= nums[i] &lt;= 104</code></li>
	<li><code>nums</code> contains distinct values sorted in ascending order.</li>
	<li><code>-104 &lt;= target &lt;= 104</code></li>
</ul>

searchInsert

Search Insert Position

First Bad Version

first-bad-version

第一个错误的版本

Minimum Operations to Exceed Threshold Value I

minimum-operations-to-exceed-threshold-value-i

超过阈值的最少操作数 I

Array

Binary Search

给定一个排序数组和一个目标值，在数组中找到目标值，并返回其索引。如果目标值不存在于数组中，返回它将会被按顺序插入的位置。

请必须使用时间复杂度为 <code>O(log n)</code> 的算法。

&nbsp;

示例 1:

<pre>
输入: nums = [1,3,5,6], target = 5
输出: 2
</pre>

示例&nbsp;2:

<pre>
输入: nums = [1,3,5,6], target = 2
输出: 1
</pre>

示例 3:

<pre>
输入: nums = [1,3,5,6], target = 7
输出: 4
</pre>

&nbsp;

提示:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 104</code></li>
	<li><code>-104 &lt;= nums[i] &lt;= 104</code></li>
	<li><code>nums</code> 为&nbsp;无重复元素&nbsp;的&nbsp;升序&nbsp;排列数组</li>
	<li><code>-104 &lt;= target &lt;= 104</code></li>
</ul>