leetcode11

public class Solution
    {
        //public int MaxArea(int[] height)
        //{
        //    var max = 0;
        //    for (int i = 0; i < height.Length; i++)
        //    {
        //        for (int j = 0; j < height.Length; j++)
        //        {
        //            if (i == j)
        //            {
        //                continue;
        //            }
        //            var min = Math.Min(height[i], height[j]);
        //            var dif = Math.Abs(i - j);
        //            var product = min * dif;
        //            max = Math.Max(max, product);
        //        }
        //    }
        //    return max;
        //}
        public int MaxArea(int[] height)
        {
            int max = int.MinValue;
            for (int i = , j = height.Length - ; i < j;)
            {
                if (height[i] > height[j])
                {
                    max = Math.Max(max, height[j] * (j - i));
                    j--;
                }
                else
                {
                    max = Math.Max(max, height[i] * (j - i));
                    i++;
                }
            }
            return max;
        }
    }

下面这个解决方案算是一种搞笑吧，夸张的6796ms，居然也能AC！

 class Solution:
     def findTopTwoNum(self,height):
         top1 = -1
         top1_index = -1

         top2 = -1
         top2_index = -1
         for i in range(len(height)):
             if top1 < height[i]:
                 top1 = height[i]
                 top1_index = i

         for i in range(len(height)):
             if i != top1_index and top2 < height[i]:
                 top2 = height[i]
                 top2_index = i

         return top1_index,top2_index 

     def maxArea(self, height: 'List[int]') -> 'int':
         #在height中找最大的两个数字，以及这两个数字的坐标
         left,right = self.findTopTwoNum(height)

         ma = min(left,right) * (right - left)
         for i in range(left,-1,-1):
             for j in range(right,len(height)):
                 area = min(height[i],height[j]) * (j-i)
                 if area > ma :
                     ma = area
         return ma