leetcode-11 Container With Most Water

/**
* 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
*
*/
public class ContainerWithMostWater {

/**
* 由题意可知即求出两个y坐标与x坐标围绕的最大面积（x之间的距离*y上的第二个最大值）
* @param height
* @return
*/
public int maxArea(int[] height) {
   int max = 0;
   if(height == null || height.length <=1) {
       return max;
   }else {
       for(int i=0;i<height.length-1;i++) {
           int tmpMax = height[i];
           int tmpHeight = tmpMax;
           for(int j=i+1;j<height.length;j++) {
               tmpHeight = tmpMax>=height[j]? height[j]:tmpMax;
               max = max > (j-i)*tmpHeight? max:(j-i)*tmpHeight;
           }
       }
   }

   return max;
}

   public static void main(String[] args) {
       // TODO Auto-generated method stub
       ContainerWithMostWater a = new ContainerWithMostWater();
       int[] as = new int[] {1,8,6,2,5,4,8,3,7};
       System.out.println(a.maxArea(as));
   }

}

以上为最麻烦的暴力的遍历

可以调用方法从两端开始 i=0 j=height.length()-1

计算出当前的值然后根据height[i]与height[j]的大小比较来判断是i++或是j--

知道i<j不成立

leetcode-11 Container With Most Water

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