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: [1,8,6,2,5,4,8,3,7] Output: 49
开始把这题想复杂了，决定最大值的只有两侧的高度，遍历的话 o(n^2)的复杂度
很多状态是重复的
可以通过双指针转化为贪心问题
对于
100001
目前的状态下移动挡板只有移动矮的挡板才能可能比当前回大
public int maxArea(int[] height) {
int max=0;
int l=0;
int r=height.length-1;
while(l<r){max=Math.max(max,Math.min(height[l],height[r])*(r-l));
if(height[l]<height[r]) l++;
else r--;
}
return max; }