From 37d963ce4f75b38eb8f05550204921c84981bf54 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Fri, 13 Aug 2021 16:50:26 +0800
Subject: [PATCH] =?UTF-8?q?713:=E4=B9=98=E7=A7=AF=E5=B0=8F=E4=BA=8EK?=
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---
 .../editor/cn/SubarrayProductLessThanK.java   | 75 +++++++++++++++++++
 .../editor/cn/SubarrayProductLessThanK.md     | 31 ++++++++
 2 files changed, 106 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.java
 create mode 100644 src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.md

diff --git a/src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.java b/src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.java
new file mode 100644
index 0000000..d5b0ebd
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.java
@@ -0,0 +1,75 @@
+//给定一个正整数数组 nums和整数 k 。 
+//
+// 请找出该数组内乘积小于 k 的连续的子数组的个数。 
+//
+// 
+//
+// 示例 1: 
+//
+// 
+//输入: nums = [10,5,2,6], k = 100
+//输出: 8
+//解释: 8个乘积小于100的子数组分别为: [10], [5], [2], [6], [10,5], [5,2], [2,6], [5,2,6]。
+//需要注意的是 [10,5,2] 并不是乘积小于100的子数组。
+// 
+//
+// 示例 2: 
+//
+// 
+//输入: nums = [1,2,3], k = 0
+//输出: 0 
+//
+// 
+//
+// 提示: 
+//
+// 
+// 1 <= nums.length <= 3 * 104 
+// 1 <= nums[i] <= 1000 
+// 0 <= k <= 106 
+// 
+// Related Topics 数组 滑动窗口 
+// 👍 272 👎 0
+
+package leetcode.editor.cn;
+
+//713:乘积小于K的子数组
+class SubarrayProductLessThanK {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new SubarrayProductLessThanK().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int numSubarrayProductLessThanK(int[] nums, int k) {
+            if (k == 0) {
+                return 0;
+            }
+            double logk = Math.log(k);
+            double[] logs = new double[nums.length + 1];
+            for (int i = 0; i < nums.length; i++) {
+                logs[i + 1] = logs[i] + Math.log(nums[i]);
+            }
+
+            int result = 0;
+            for (int i = 0; i < logs.length; i++) {
+                int start = i + 1;
+                int end = logs.length;
+                while (start < end) {
+                    int mid = (end + start) / 2;
+                    if (logs[mid] < logs[i] + logk - 1e-9) {
+                        start = mid + 1;
+                    } else {
+                        end = mid;
+                    }
+                }
+                result += start - i - 1;
+            }
+            return result;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.md b/src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.md
new file mode 100644
index 0000000..0ec7a1c
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/SubarrayProductLessThanK.md
@@ -0,0 +1,31 @@
+<p>给定一个正整数数组 <code>nums</code>和整数 <code>k</code> 。</p>
+
+<p>请找出该数组内乘积小于 <code>k</code> 的连续的子数组的个数。</p>
+
+<p> </p>
+
+<p><strong>示例 1:</strong></p>
+
+<pre>
+<strong>输入:</strong> nums = [10,5,2,6], k = 100
+<strong>输出:</strong> 8
+<strong>解释:</strong> 8个乘积小于100的子数组分别为: [10], [5], [2], [6], [10,5], [5,2], [2,6], [5,2,6]。
+需要注意的是 [10,5,2] 并不是乘积小于100的子数组。
+</pre>
+
+<p><strong>示例 2:</strong></p>
+
+<pre>
+<strong>输入:</strong> nums = [1,2,3], k = 0
+<strong>输出:</strong> 0</pre>
+
+<p> </p>
+
+<p><strong>提示: </strong></p>
+
+<ul>
+	<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>
+	<li><code>1 <= nums[i] <= 1000</code></li>
+	<li><code>0 <= k <= 10<sup>6</sup></code></li>
+</ul>
+<div><div>Related Topics</div><div><li>数组</li><li>滑动窗口</li></div></div>\n<div><li>👍 272</li><li>👎 0</li></div>
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