From ed660c8b5760a2b515ab7305a9e3680916c56d3b Mon Sep 17 00:00:00 2001
From: "huangge1199@hotmail.com" <huangge1199>
Date: Tue, 6 Jul 2021 23:40:38 +0800
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 .../leetcode/editor/cn/MaximumSubarray.java   | 79 +++++++++++++++++++
 .../leetcode/editor/cn/MaximumSubarray.md     | 53 +++++++++++++
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 create mode 100644 src/main/java/leetcode/editor/cn/MaximumSubarray.java
 create mode 100644 src/main/java/leetcode/editor/cn/MaximumSubarray.md

diff --git a/src/main/java/leetcode/editor/cn/MaximumSubarray.java b/src/main/java/leetcode/editor/cn/MaximumSubarray.java
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+//给定一个整数数组 nums ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：nums = [-2,1,-3,4,-1,2,1,-5,4]
+//输出：6
+//解释：连续子数组 [4,-1,2,1] 的和最大，为 6 。
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：nums = [1]
+//输出：1
+// 
+//
+// 示例 3： 
+//
+// 
+//输入：nums = [0]
+//输出：0
+// 
+//
+// 示例 4： 
+//
+// 
+//输入：nums = [-1]
+//输出：-1
+// 
+//
+// 示例 5： 
+//
+// 
+//输入：nums = [-100000]
+//输出：-100000
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= nums.length <= 3 * 104 
+// -105 <= nums[i] <= 105 
+// 
+//
+// 
+//
+// 进阶：如果你已经实现复杂度为 O(n) 的解法，尝试使用更为精妙的 分治法 求解。 
+// Related Topics 数组 分治 动态规划 
+// 👍 3368 👎 0
+
+package leetcode.editor.cn;
+
+//53:最大子序和
+class MaximumSubarray {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new MaximumSubarray().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int maxSubArray(int[] nums) {
+            int pre = 0, maxAns = nums[0];
+            for (int x : nums) {
+                pre = Math.max(pre + x, x);
+                maxAns = Math.max(maxAns, pre);
+            }
+            return maxAns;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/MaximumSubarray.md b/src/main/java/leetcode/editor/cn/MaximumSubarray.md
new file mode 100644
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+++ b/src/main/java/leetcode/editor/cn/MaximumSubarray.md
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+<p>给定一个整数数组 <code>nums</code> ，找到一个具有最大和的连续子数组（子数组最少包含一个元素），返回其最大和。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [-2,1,-3,4,-1,2,1,-5,4]
+<strong>输出：</strong>6
+<strong>解释：</strong>连续子数组 [4,-1,2,1] 的和最大，为 6 。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [1]
+<strong>输出：</strong>1
+</pre>
+
+<p><strong>示例 3：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [0]
+<strong>输出：</strong>0
+</pre>
+
+<p><strong>示例 4：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [-1]
+<strong>输出：</strong>-1
+</pre>
+
+<p><strong>示例 5：</strong></p>
+
+<pre>
+<strong>输入：</strong>nums = [-100000]
+<strong>输出：</strong>-100000
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>
+	<li><code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code></li>
+</ul>
+
+<p> </p>
+
+<p><strong>进阶：</strong>如果你已经实现复杂度为 <code>O(n)</code> 的解法，尝试使用更为精妙的 <strong>分治法</strong> 求解。</p>
+<div><div>Related Topics</div><div><li>数组</li><li>分治</li><li>动态规划</li></div></div>\n<div><li>👍 3368</li><li>👎 0</li></div>
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