From a854f1c12d174d3d52e26649f9ca39112c9b6ae4 Mon Sep 17 00:00:00 2001
From: zeekling <lingzhaohui@zeekling.cn>
Date: Sun, 28 Feb 2021 23:16:52 +0800
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---
 .../java/com/leetcode/huawei/MaxSubArray.java | 33 ++++++++++++++++++
 .../huawei/MinCostClimbingStairs.java         | 34 +++++++++++++++++++
 2 files changed, 67 insertions(+)
 create mode 100644 src/main/java/com/leetcode/huawei/MaxSubArray.java
 create mode 100644 src/main/java/com/leetcode/huawei/MinCostClimbingStairs.java

diff --git a/src/main/java/com/leetcode/huawei/MaxSubArray.java b/src/main/java/com/leetcode/huawei/MaxSubArray.java
new file mode 100644
index 0000000..28af6d8
--- /dev/null
+++ b/src/main/java/com/leetcode/huawei/MaxSubArray.java
@@ -0,0 +1,33 @@
+package com.leetcode.huawei;
+
+/**
+ * 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。
+ * 
+ * https://leetcode-cn.com/problems/maximum-subarray/
+ */
+public class MaxSubArray {
+
+
+    public int maxSubArray(int[] nums) {
+        int res = nums[0];
+        int sum = 0;
+        for (int num : nums) {
+            if (sum > 0)
+                sum += num;
+            else
+                sum = num;
+            res = Math.max(res, sum);
+        }
+        return res;
+    }
+
+
+
+    public static void main(String[] args) {
+        MaxSubArray maArray = new MaxSubArray();
+        int[] arr = {-2,1,-3,4,-1,2,1,-5,4};
+        int result = maArray.maxSubArray(arr);
+        System.out.println(result);
+    }
+    
+}
diff --git a/src/main/java/com/leetcode/huawei/MinCostClimbingStairs.java b/src/main/java/com/leetcode/huawei/MinCostClimbingStairs.java
new file mode 100644
index 0000000..c2e0cae
--- /dev/null
+++ b/src/main/java/com/leetcode/huawei/MinCostClimbingStairs.java
@@ -0,0 +1,34 @@
+package com.leetcode.huawei;
+/**
+ * 
+ 数组的每个下标作为一个阶梯，第 i 个阶梯对应着一个非负数的体力花费值 cost[i]（下标从 0 开始）。
+每当你爬上一个阶梯你都要花费对应的体力值，一旦支付了相应的体力值，你就可以选择向上爬一个阶梯或者爬两个阶梯。
+请你找出达到楼层顶部的最低花费。在开始时，你可以选择从下标为 0 或 1 的元素作为初始阶梯。
+示例 1：
+
+输入：cost = [10, 15, 20]
+输出：15
+解释：最低花费是从 cost[1] 开始，然后走两步即可到阶梯顶，一共花费 15 。
+ 示例 2：
+
+输入：cost = [1, 100, 1, 1, 1, 100, 1, 1, 100, 1]
+输出：6
+解释：最低花费方式是从 cost[0] 开始，逐个经过那些 1 ，跳过 cost[3] ，一共花费 6 。
+
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/min-cost-climbing-stairs
+著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+ * 
+ */
+public class MinCostClimbingStairs {
+    
+    public int minCostClimbingStairs(int[] cost) {
+
+        return 0;
+    }
+
+
+    public static void main(String[] args) {
+        
+    }
+}