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From: huangge1199 <huangge1199@hotmail.com>
Date: Fri, 13 Aug 2021 16:24:32 +0800
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 .../editor/cn/MinimumSizeSubarraySum.java     | 81 +++++++++++++++++++
 .../editor/cn/MinimumSizeSubarraySum.md       | 46 +++++++++++
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 create mode 100644 src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.java
 create mode 100644 src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.md

diff --git a/src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.java b/src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.java
new file mode 100644
index 0000000..627029d
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+++ b/src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.java
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+//给定一个含有 n 个正整数的数组和一个正整数 target 。 
+//
+// 找出该数组中满足其和 ≥ target 的长度最小的 连续子数组 [numsl, numsl+1, ..., numsr-1, numsr] ，并返回其长
+//度。如果不存在符合条件的子数组，返回 0 。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：target = 7, nums = [2,3,1,2,4,3]
+//输出：2
+//解释：子数组 [4,3] 是该条件下的长度最小的子数组。
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：target = 4, nums = [1,4,4]
+//输出：1
+// 
+//
+// 示例 3： 
+//
+// 
+//输入：target = 11, nums = [1,1,1,1,1,1,1,1]
+//输出：0
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= target <= 109 
+// 1 <= nums.length <= 105 
+// 1 <= nums[i] <= 105 
+// 
+//
+// 
+//
+// 进阶： 
+//
+// 
+// 如果你已经实现 O(n) 时间复杂度的解法, 请尝试设计一个 O(n log(n)) 时间复杂度的解法。 
+// 
+// Related Topics 数组 二分查找 前缀和 滑动窗口 
+// 👍 723 👎 0
+
+package leetcode.editor.cn;
+
+//209:长度最小的子数组
+class MinimumSizeSubarraySum {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new MinimumSizeSubarraySum().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int minSubArrayLen(int target, int[] nums) {
+            int sum = 0;
+            int start = 0;
+            int min = Integer.MAX_VALUE;
+            for (int i = 0; i < nums.length; i++) {
+                sum += nums[i];
+                if (sum >= target) {
+                    while (sum >= target) {
+                        sum -= nums[start];
+                        start++;
+                    }
+                    min = Math.min(min, i - start + 2);
+                }
+            }
+            return min == Integer.MAX_VALUE ? 0 : min;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.md b/src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.md
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index 0000000..be18656
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+++ b/src/main/java/leetcode/editor/cn/MinimumSizeSubarraySum.md
@@ -0,0 +1,46 @@
+<p>给定一个含有 <code>n</code><strong> </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和<strong> </strong><code>≥ target</code><strong> </strong>的长度最小的 <strong>连续子数组</strong> <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> </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 <code>[4,3]</code> 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> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= target <= 10<sup>9</sup></code></li>
+	<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
+	<li><code>1 <= nums[i] <= 10<sup>5</sup></code></li>
+</ul>
+
+<p> </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>
+<div><div>Related Topics</div><div><li>数组</li><li>二分查找</li><li>前缀和</li><li>滑动窗口</li></div></div>\n<div><li>👍 723</li><li>👎 0</li></div>
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