From 5926432b0ace0537b6dabaa154a48b44b0aae06d Mon Sep 17 00:00:00 2001
From: whai <lkwhai@163.com>
Date: Sun, 10 Dec 2023 21:09:11 +0800
Subject: [PATCH] =?UTF-8?q?LeetCode209=20=E5=8F=8C=E6=8C=87=E9=92=88?=
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---
 .../whaifree/leetCode/middle/LeetCode209.java | 90 +++++++++++++++++++
 1 file changed, 90 insertions(+)
 create mode 100644 src/main/java/cn/whaifree/leetCode/middle/LeetCode209.java

diff --git a/src/main/java/cn/whaifree/leetCode/middle/LeetCode209.java b/src/main/java/cn/whaifree/leetCode/middle/LeetCode209.java
new file mode 100644
index 0000000..838a30f
--- /dev/null
+++ b/src/main/java/cn/whaifree/leetCode/middle/LeetCode209.java
@@ -0,0 +1,90 @@
+package cn.whaifree.leetCode.middle;
+
+import org.junit.Test;
+
+/**
+ * 209. 长度最小的子数组
+ * 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总和大于等于 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。
+ */
+public class LeetCode209 {
+
+    /**
+     * 暴力求解，会超时
+     * @param target
+     * @param nums
+     * @return
+     */
+    public int minSubArrayLen(int target, int[] nums) {
+
+        int minLength = Integer.MAX_VALUE;
+        for (int i = 0; i < nums.length; i++) {
+            int sum = 0;
+            for (int j = i; j < nums.length; j++) {
+                sum += nums[j];
+                if (sum >= target) {
+                    minLength = Math.min(minLength, j - i + 1);
+                    break;
+                }
+                /**
+                 * if (sum >= target && j - i +1 < minLength) {
+                 *                     minLength = j - i + 1;
+                 *                     break;
+                 *                 }
+                 */
+            }
+        }
+
+
+        return minLength == Integer.MAX_VALUE ? 0 : minLength;
+    }
+
+    public int minSubArrayLen1(int target, int[] nums) {
+        int left = 0;
+        int right = 0;
+        int sum = 0;
+        int ans = Integer.MAX_VALUE;
+        while (right < nums.length ) {
+            sum += nums[right];
+            // 窗口内，找到最小子串
+            while (sum >= target) {
+                ans = Math.min(ans, right - left + 1);
+                sum -= nums[left++];
+            }
+            right++;
+        }
+        return ans == Integer.MAX_VALUE ? 0 : ans;
+    }
+
+
+    @Test
+    public void list() {
+        System.out.println(minSubArrayLen1(5, new int[]{2,3,6}));
+    }
+
+}