Container with the most water

Here you are. n Nonnegative integers a1,a2,…,an, Each number represents a point in the coordinate (i, ai) . Draw in coordinates n Bar vertical line , Vertical line i The two endpoints of are (i, ai) and (i, 0) . Find two of them , Make them x A container of shafts can hold the most water .

explain ： You can't tilt the container .

example 1

Input ：[1,8,6,2,5,4,8,3,7]
Output ：49
explain ： The vertical line in the figure represents the input array [1,8,6,2,5,4,8,3,7]. In this case , The container can hold water （ In blue ） The maximum value of 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

Method 1 （ I failed to pass this method because I exceeded the time limit during the test ）

public class Solution {
    
    public int maxArea(int[] height) {
    
        int max = 0;
        for (int i = 0; i < height.length - 1; ++i) {
    
            for (int j = i + 1; j < height.length; ++j) {
    
                int area = (j - i)*Math.min(height[i],height[j]);
                max = Math.max(area,max);
            }
        }
        return max; 
    }
}

Method 2 （ Be careful i++ It's assignment before operation ）

public class Solution {
    
    public int maxArea(int[] a) {
    
        int max = 0;
        for (int i = 0,j = a.length - 1; i < j;) {
    
            int minHeight = a[i] < a[j] ? a[i++] : a[j--];
            int area = (j - i+1)*minHeight;// because i++ and i-- They are all assignment first and then operation , So we need to add one more 1 To calculate accurately 
            max = Math.max(max,area);
        }
        return max;
    }
}

Method 3

class Solution {
    
    public int maxArea(int[] a) {
    
        int i = 0,j= a.length - 1,res = 0;
        while (i < j) {
    
            res = a[i] < a[j] ?
                    Math.max(res,(j - i)*a[i++])://i++ and i-- They are all assignment first and then operation , At this time or i
                    Math.max(res,(j - i)*a[i--]);
        }
        return res;
    }
}