feat: add 011

Author: Blankj

README.md

@@ -72,6 +72,7 @@
 | 5    | [Longest Palindromic Substring][005]     | String                           |
 | 6    | [ZigZag Conversion][006]                 | String                           |
 | 8    | [String to Integer (atoi)][008]          | Math, String                     |
+| 11   | [Container With Most Water][011]         | Array, Two Pointers              |
 | 15   | [3Sum][015]                              | Array, Two Pointers              |
 | 17   | [Letter Combinations of a Phone Number][017] | String, Backtracking             |
 | 19   | [Remove Nth Node From End of List][019]  | Linked List, Two Pointers        |
@@ -141,6 +142,7 @@
 [005]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/005/README.md
 [006]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/006/README.md
 [008]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/008/README.md
+[011]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/011/README.md
 [015]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/015/README.md
 [017]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/017/README.md
 [019]: https://github.com/Blankj/awesome-java-leetcode/blob/master/note/019/README.md

note/011/README.md

@@ -11,12 +11,30 @@ Note: You may not slant the container and *n* is at least 2.
 
 ## 思路
 
+题意是给你 *a1*, *a2*, ..., *an* 这 *n* 个数，代表 (*i*, *ai*) 坐标，让你从中找两个点与 x 轴围成的容器可以容纳最多的水。
 
+不明白的话可以看数据为 `1 8 6 2 5 4 8 3 7` 所示的图。
 
-```java
+![](https://raw.githubusercontent.com/Blankj/awesome-java-leetcode/master/note/011/water.png)
+
+如果用暴力法求每种情况的结果，其时间复杂度为 O(n^2)，相信肯定会超时，我们可以探索下是否有更巧妙的办法呢，题目的标签有双指针，是否就可以想到首尾各放一指针，然后根据条件来收缩。首先计算一次首尾构成的最大面积，然后分析下该移动哪个指针，如果移动大的那个指针的话，那样只会减小面积，所以我们要移动小的那个指针，小的那个指针移动到哪呢？当然是移动到大于之前的值的地方，否则面积不都比之前小么，然后继续更新最大值即可，借助如上分析写出如下代码应该不是什么难事了吧。
 
-```
 
+```java
+class Solution {
+    public int maxArea(int[] height) {
+        int l = 0, r = height.length - 1;
+        int max = 0, h = 0;
+        while (l < r) {
+            h = Math.min(height[l], height[r]);
+            max = Math.max(max, (r - l) * h);
+            while (height[l] <= h && l < r) ++l;
+            while (height[r] <= h && l < r) --r;
+        }
+        return max;
+    }
+}
+```
 
 
 ## 结语

note/011/water.png

@@ -0,0 +1,29 @@
+package com.blankj.medium._011;
+
+/**
+ * <pre>
+ *     author: Blankj
+ *     blog  : http://blankj.com
+ *     time  : 2017/04/23
+ *     desc  :
+ * </pre>
+ */
+public class Solution {
+    public int maxArea(int[] height) {
+        int l = 0, r = height.length - 1;
+        int max = 0, h = 0;
+        while (l < r) {
+            h = Math.min(height[l], height[r]);
+            max = Math.max(max, (r - l) * h);
+            while (height[l] <= h && l < r) ++l;
+            while (height[r] <= h && l < r) --r;
+        }
+        return max;
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+        System.out.println(solution.maxArea(new int[]{1, 2, 4, 3})); // 4
+        System.out.println(solution.maxArea(new int[]{1, 8, 6, 2, 5, 4, 8, 3, 7}));// 49
+    }
+}