Project Notes

Minimum Size Subarray Sum

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Description

tab: English
 
<p>Given an array of positive integers <code>nums</code> and a positive integer <code>target</code>, return <em>the <strong>minimal length</strong> of a </em><span data-keyword="subarray-nonempty"><em>subarray</em></span><em> whose sum is greater than or equal to</em> <code>target</code>. If there is no such subarray, return <code>0</code> instead.</p>
 
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
 
<pre>
<strong>Input:</strong> target = 7, nums = [2,3,1,2,4,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The subarray [4,3] has the minimal length under the problem constraint.
</pre>
 
<p><strong class="example">Example 2:</strong></p>
 
<pre>
<strong>Input:</strong> target = 4, nums = [1,4,4]
<strong>Output:</strong> 1
</pre>
 
<p><strong class="example">Example 3:</strong></p>
 
<pre>
<strong>Input:</strong> target = 11, nums = [1,1,1,1,1,1,1,1]
<strong>Output:</strong> 0
</pre>
 
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>
 
<ul>
	<li><code>1 &lt;= target &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>
 
<p>&nbsp;</p>
<strong>Follow up:</strong> If you have figured out the <code>O(n)</code> solution, try coding another solution of which the time complexity is <code>O(n log(n))</code>.
 
 
---
 
[submissions](https://leetcode.com/problems/minimum-size-subarray-sum/submissions/) | [solutions](https://leetcode.com/problems/minimum-size-subarray-sum/solutions/)
 
 
tab: 中文
 
<p>给定一个含有&nbsp;<code>n</code><strong>&nbsp;</strong>个正整数的数组和一个正整数 <code>target</code><strong> 。</strong></p>
 
<p>找出该数组中满足其总和大于等于<strong> </strong><code>target</code><strong> </strong>的长度最小的 <strong><span data-keyword="subarray-nonempty">子数组</span></strong>&nbsp;<code>[nums<sub>l</sub>, nums<sub>l+1</sub>, ..., nums<sub>r-1</sub>, nums<sub>r</sub>]</code> ，并返回其长度<strong>。</strong>如果不存在符合条件的子数组，返回 <code>0</code> 。</p>
 
<p>&nbsp;</p>
 
<p><strong>示例 1：</strong></p>
 
<pre>
<strong>输入：</strong>target = 7, nums = [2,3,1,2,4,3]
<strong>输出：</strong>2
<strong>解释：</strong>子数组&nbsp;<code>[4,3]</code>&nbsp;是该条件下的长度最小的子数组。
</pre>
 
<p><strong>示例 2：</strong></p>
 
<pre>
<strong>输入：</strong>target = 4, nums = [1,4,4]
<strong>输出：</strong>1
</pre>
 
<p><strong>示例 3：</strong></p>
 
<pre>
<strong>输入：</strong>target = 11, nums = [1,1,1,1,1,1,1,1]
<strong>输出：</strong>0
</pre>
 
<p>&nbsp;</p>
 
<p><strong>提示：</strong></p>
 
<ul>
	<li><code>1 &lt;= target &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>
 
<p>&nbsp;</p>
 
<p><strong>进阶：</strong></p>
 
<ul>
	<li>如果你已经实现<em> </em><code>O(n)</code> 时间复杂度的解法, 请尝试设计一个 <code>O(n log(n))</code> 时间复杂度的解法。</li>
</ul>
 
 
 
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[提交记录](https://leetcode.cn/problems/minimum-size-subarray-sum/submissions/) | [题解](https://leetcode.cn/problems/minimum-size-subarray-sum/solution/)

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