# # 11. Container With Most Water

個人最快的解法是用2個指標，一個從最左，一個從最右。左一右就是長度。面積就是長度\*min(\[左]，\[右])，最次面積就存到comp，再跟max2比大小，comp比max2大就把comp指定給max2。

並判斷最左的值與最右的值大小，左比右小就把左向右移動一格; 若右比左小就把右向左移動一格(動的都是低的那根)，以確保目前的長度的是最高的。

```java
class Solution {
    public int maxArea(int[] height) {
        //two number ,the y is the shower one , power 2
        //two number, the x is the one number and the two number index.
        
        int font = 0;
        int end = height.length-1;
        int max2 = 0;
        int comp = 0;
        while(font < end){
            comp = (end - font) *findMin( height[font] ,height[end] );
            if(comp > max2){
                max2= comp;
            }
            // System.out.println("font = " + font + " , end = " + end + " [f] = " + height[font] + " , [e] = " + height[end] + " ,comp = " + comp );
            if(height[font] > height[end] ){
                end--;
            }else{
                font++;
            }
        }
        return max2;
        //fast 27% 
//         if(height.length == 2){
//             return 1* findMin( height[0] , height[1] );
//         }
//         int max2 = 0;
//         int comp = 0;
//         int cout = 0;
//        for(int x = 0 ; x < height.length; x++){
//            if(x > 0){
//                cout = x;
//                for(int y = 0 ; y < x ; y++){
//                    comp = cout * findMin( height[x] , height[y] );
//                    // System.out.println("x = " + x + " , h["+x+"]= "+ height[x] + " , y = " + y + " , h["+y+"]= "+ height[y] + " ,the comp = " + comp + " ,cout = " + cout );
//                    if(comp > max2 ){
//                        max2 = comp;
//                    }
//                    // System.out.println("--------------------------\n");
//                    cout--;
//                }
//            }
//        }
        
//         return max2;
    }
    
    public int findMin(int a, int b){
        if(a > b){
            return b;
        }
        return a;
    }
}
```
