finish "152. Maximum Product Subarray". Fix #152

Author: diguage

Diff for: README.adoc

@@ -763,12 +763,12 @@
 //|{leetcode_base_url}/reverse-words-in-a-string/[Reverse Words in a String]
 //|{source_base_url}/ReverseWordsInAString.java[Java]
 //|Medium
-//
-//|152
-//|{leetcode_base_url}/maximum-product-subarray/[Maximum Product Subarray]
-//|{source_base_url}/MaximumProductSubarray.java[Java]
-//|Medium
-//
+
+|152
+|{leetcode_base_url}/maximum-product-subarray/[Maximum Product Subarray]
+|{source_base_url}/MaximumProductSubarray.java[Java]
+|Medium
+
 //|153
 //|{leetcode_base_url}/find-minimum-in-rotated-sorted-array/[Find Minimum in Rotated Sorted Array]
 //|{source_base_url}/FindMinimumInRotatedSortedArray.java[Java]

Diff for: docs/0152-maximum-product-subarray.adoc

@@ -0,0 +1,4 @@
+== 152. Maximum Product Subarray
+
+这道题的阶梯思路跟另外某个题的解题思路很类似！
+

Diff for: src/main/java/com/diguage/algorithm/leetcode/MaximumProductSubarray.java

@@ -0,0 +1,71 @@
+package com.diguage.algorithm.leetcode;
+
+/**
+ * = 152. Maximum Product Subarray
+ *
+ * https://leetcode.com/problems/maximum-product-subarray/[Maximum Product Subarray - LeetCode]
+ *
+ * Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product.
+ *
+ * .Example 1:
+ * [source]
+ * ----
+ * Input: [2,3,-2,4]
+ * Output: 6
+ * Explanation: [2,3] has the largest product 6.
+ * ----
+ *
+ * .Example 2:
+ * [source]
+ * ----
+ * Input: [-2,0,-1]
+ * Output: 0
+ * Explanation: The result cannot be 2, because [-2,-1] is not a subarray.
+ * ----
+ *
+ * @author D瓜哥, https://www.diguage.com/
+ * @since 2020-01-05 07:59
+ */
+public class MaximumProductSubarray {
+    /**
+     * Runtime: 1 ms, faster than 97.91% of Java online submissions for Maximum Product Subarray.
+     *
+     * Memory Usage: 36.4 MB, less than 100.00% of Java online submissions for Maximum Product Subarray.
+     *
+     * Copy form: https://leetcode.com/problems/maximum-product-subarray/discuss/48230/Possibly-simplest-solution-with-O(n)-time-complexity[Possibly simplest solution with O(n) time complexity - LeetCode Discuss]
+     */
+    public int maxProduct(int[] nums) {
+        // store the result that is the max we have found so far
+        int result = nums[0];
+        // max/min stores the max/min product of
+        // subarray that ends with the current number nums[i]
+        for (int i = 1, max = result, min = result; i < nums.length; i++) {
+            // multiplied by a negative makes big number smaller, small number bigger
+            // so we redefine the extremums by swapping them
+            if (nums[i] < 0) {
+                int temp = max;
+                max = min;
+                min = temp;
+            }
+            // max/min product for the current number is either the current number itself
+            // or the max/min by the previous number times the current one
+            min = Math.min(nums[i], min * nums[i]);
+            max = Math.max(nums[i], max * nums[i]);
+
+            // the newly computed max value is a candidate for our global result
+            result = Math.max(result, max);
+        }
+        return result;
+    }
+
+    public static void main(String[] args) {
+        MaximumProductSubarray solution = new MaximumProductSubarray();
+        int[] n1 = {2, 3, -2, 4};
+        int r1 = solution.maxProduct(n1);
+        System.out.println((r1 == 6) + " : " + r1);
+
+        int[] n2 = {-2, 0, -1};
+        int r2 = solution.maxProduct(n2);
+        System.out.println((r2 == 0) + " : " + r1);
+    }
+}