From a952faff55a36999dc5cb283b4df6d952b864914 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?=E5=94=90=E6=A0=91=E6=A3=AE?= <14021051@buaa.edu.cn>
Date: Sat, 8 Dec 2018 12:41:48 +0800
Subject: [PATCH] Create 11. Container With Most Water.md

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+# [11. Container With Most Water](https://leetcode.com/problems/container-with-most-water/)
+# 思路
+题意就是选择两条线，使其组成的容器装的水最多。水的量可以用宽乘高计算。   
+由于宽最大就是height两端之间的距离，所以要想在两端之内取得最大值的话只能是高度比较高。   
+可以考虑设置两个初始分别为两端的指针left和right代表当前容器，不断跳过高度不够高的height使这两个指针往中间靠。这个过程不断循环即可得到结果。   
+时间复杂度O(n)，空间复杂度O(1)
+
+# C++
+``` C++
+class Solution {
+public:
+    int maxArea(vector<int>& height) {
+        int res = 0;
+        int left = 0, right = height.size() - 1;
+        while(left < right){
+            int h = min(height[left], height[right]); // 当前容器高度
+            res = max(res, h * (right - left));
+            while(height[left] <= h) left++;  // 跳过高度不够高的
+            while(height[right] <= h) right--;
+        }
+        return res;
+    }
+};
+```