【题目】

木桶短板，能装的最多水，由最短的那条决定

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 the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

Notice that you may not slant the container.

Example 1:

Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: 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 2:

Input: height = [1,1]
Output: 1

Example 3:

Input: height = [4,3,2,1,4]
Output: 16

Example 4:

Input: height = [1,2,1]
Output: 2

 【思路】

木桶短板，能装的最多水，由最短的那条决定

头尾俩指针，计算面积max更新

判断左右哪边更小（木桶更短板），更矮的向对面移动，left<right

因为小的移动，下一个值可能更大，面积也可能更大

【代码】

class Solution {
    public int maxArea(int[] height) {
        int area=0;
        int left=0;
        int right=height.length-1;
        while(left<right){
            area=Math.max(area,Math.min(height[left],height[right])*(right-left));
            if(height[left]<height[right])
                left++;
            else
                right--;
        }
        return area;
    }
}