# 209. Minimum Size Subarray Sum

Given an array of positive integers `nums` and a positive integer `target`, return *the **minimal length** of a subarray whose sum is greater than or equal to* `target`. If there is no such subarray, return `0` instead.

**Example 1:**

<pre><code><strong>Input: target = 7, nums = [2,3,1,2,4,3]
</strong><strong>Output: 2
</strong><strong>Explanation: The subarray [4,3] has the minimal length under the problem constraint.
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: target = 4, nums = [1,4,4]
</strong><strong>Output: 1
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: target = 11, nums = [1,1,1,1,1,1,1,1]
</strong><strong>Output: 0
</strong></code></pre>

***My Solutions:*** 这道题和76相似。用sliding window记录大于等于target的范围，如果超过了target，可以把sliding window的左边向右移动。

Time: O(n); Space: O(1)

<pre class="language-java"><code class="lang-java"><strong>public int minSubArrayLen(int target, int[] nums) {
</strong>    int min = nums.length + 1, l = 0, sum = 0;
    for (int i = 0; i &#x3C; nums.length; i++) {
        sum += nums[i];
        while (sum >= target) {
            min = Math.min(min, i - l + 1);
            // remove the first element from the window
            sum -= nums[l];
            // move the start of window --> left
            l++;
        }
    }
    return min > nums.length ? 0 : min;
}
</code></pre>
