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.
The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
Example:
Input: [,,,,,,,,]
Output:
题目描述：给定n（n>=2）个非负整数，构成下面的图，找到两条线，它们与x轴一起形成一个容器，这样容器就含有最多的水。
思路：可以理解为求矩形的最大面积。将y轴的点到x轴的垂直距离作为矩形的一边，两个点的x轴的值的差作为另一边。求这两个边所形成的矩形的最大面积。
两个指针分别指向给定数组的头和尾，计算面积S=(两个点的y轴的最小值)* 两个点的x轴的差值。y轴的小的点在计算之后继续向中间靠拢。直到头尾指针相邻。
代码：
//11. Container With Most Water
public int MaxArea(int[] height)
{
int maxArea = int.MinValue;
int lo = , hi = height.Length - ;
while(lo < hi)
{
int temp = ;
if(height[lo] <= height[hi])
{
temp = height[lo] * (hi - lo);
lo++;
}
else
{
temp = height[hi] * (hi - lo);
hi--;
}
if (temp > maxArea)
maxArea = temp;
}
return maxArea;
}
