第一周作业：11 container with most water

 题目：

Given n non-negative integers a₁,a₂, ..., a_n, where each represents a point at coordinate (i,a_i). n vertical lines are drawn such that the two endpoints of linei is at (i, a_i) 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.

代码：

 int maxArea(vector<int>& height) {        int water=0;        int i=0;        int j=height.size()-1;       while(i < j)        {           water = max(water, (j - i) * min(height[i], height[j]));            if(height[i] < height[j])                ++i;            else                --j;        }       return water;    }

题解：首先的思路：两个循环穷举比较求最大值，估计会超时，没试。于是设两个指针i,j分别指向头尾，并且设置个初始最大值为i,j指向头尾所形成的容积，在i<j的大前提下。比较i,j所指向的点的高度，移动i,j。如果H(i)<H(j)，则形成新的容积肯定比初始值小，所以跳过不计，如果H(i)>H(j)，则让新的容积与上一次比较，右边指针同理，直至其相遇，所求即为最大值。