CrackerCat
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diff --git a/‎src/com/blankj/medium/_1014/Solution.java
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@@ -88,6 +88,7 @@
 | 50   | [Pow(x, n)][050]                         | Math, Binary Search              |
 | 56   | [Merge Intervals][056]                   | Array, Sort                      |
 | 554  | [Brick Wall][554]                        | Hash Table                       |
+| 1014 | [最佳观光组合][1014]                       | 数组                              |
 
 
 ## Hard
@@ -170,6 +171,7 @@
 [050]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/050/README.md
 [056]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/056/README.md
 [554]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/554/README.md
+[1014]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/1014/README.md
 
 [004]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/004/README.md
 [010]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/010/README.md
 
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+# [最佳观光组合][title]
+
+## 题目描述
+
+给定正整数数组 `A`，`A[i]` 表示第 `i` 个观光景点的评分，并且两个景点 `i` 和 `j` 之间的距离为 `j - i`。
+
+一对景点（`i < j`）组成的观光组合的得分为（`A[i] + A[j] + i - j`）：景点的评分之和**减去**它们两者之间的距离。
+
+返回一对观光景点能取得的最高分。
+
+**示例：**
+
+```
+输入：[8,1,5,2,6]
+输出：11
+解释：i = 0, j = 2, A[i] + A[j] + i - j = 8 + 5 + 0 - 2 = 11
+```
+
+**提示：**
+
+1.  `2 <= A.length <= 50000`
+2.  `1 <= A[i] <= 1000`
+
+**标签：** 数组
+
+
+## 思路
+
+直接暴力两层 for 循环肯定过不了关，我们把公式变化为 `(A[i] + i) + (A[j] - j)`，看到此应该就可以想到在每次遍历 `j` 时，只需要知道 `max(A[i] + i)` 即可。
+
+```java
+class Solution {
+
+    public int maxScoreSightseeingPair(int[] A) {
+        int ans = 0, cur = A[0] + 0;
+        for (int j = 1; j < A.length; j++) {
+            ans = Math.max(ans, cur + A[j] - j); // 计算当前最大得分
+            cur = Math.max(cur, A[j] + j);       // 更新最大的 A[i] + i
+        }
+        return ans;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        int[] A = new int[]{8, 1, 5, 2, 6};
+        System.out.println(solution.maxScoreSightseeingPair(A));
+    }
+}
+
+```
+
+
+## 结语
+
+如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]
+
+
+
+[title]: https://leetcode-cn.com/problems/best-sightseeing-pair/
+[ajl]: https://github.com/Blankj/awesome-java-leetcode
@@ -0,0 +1,27 @@
+package com.blankj.medium._1014;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2020/06/18
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+
+    public int maxScoreSightseeingPair(int[] A) {
+        int ans = 0, cur = A[0] + 0;
+        for (int j = 1; j < A.length; j++) {
+            ans = Math.max(ans, cur + A[j] - j); // 计算当前最大得分
+            cur = Math.max(cur, A[j] + j);       // 更新最大的 A[i] + i
+        }
+        return ans;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        int[] A = new int[]{8, 1, 5, 2, 6};
+        System.out.println(solution.maxScoreSightseeingPair(A));
+    }
+}