11. Container With Most Water

题面

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

给定n个非负数字，代表n条垂线，在垂线之间装水，找出能装水最多的情况。

样例（如上图）

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49
statement: (8-1)*min(8, 7)

思路1：

暴力法：遍历所有可能情况，找出最大值。

时间复杂度：O(n2)

空间复杂度：O(1)

 1 class Solution {
 2 public:
 3     int maxArea(vector<int>& height) {
 4         int size = height.size();
 5         if(size == 2)
 6             return min(height[0], height[1]);
 7         int res = 0;
 8         for(int i=0; i<size; i++)
 9         {
10             for(int j=i+1; j<size; j++)
11             {
12                 int tmp = (j-i)*min(height[j],height[i]);
13                 if(tmp > res)
14                     res = tmp;
15             }
16         }
17         return res;
18     }
19 };

思路2：

两点逼近思想

1. 先取最左端和和最右端，然后计算容积，更新容积；

2. 更新l and r, 那一端长度height[]短，那一段就更新（++/--）；

3. 输出最大容积。

时间复杂度：O(n)

空间复杂度：O(1)

 1 class Solution {
 2 public:
 3     int maxArea(vector<int>& height) {
 4         int size = height.size();
 5         if(size == 2)
 6             return min(height[0], height[1]);
 7         int res = 0;
 8         int l =0, r = size-1;
 9         while(l < r)
10         {
11             int tmp = (r-l)*min(height[r],height[l]);
12             if(tmp > res)
13                 res = tmp;
14             if(height[l] > height[r])
15                 r--;
16             else
17                 l++;
18         }
19         return res;
20     }
21 };