11 盛水最多的容器

每每看及官方解法,总羞愧难当,官方解法简洁易懂,何时才能达到这种水平呢?只有不断努力了。

package leetCode;
//官方解法
class Solution {
  public int maxArea(int[] height) {
      int maxarea = 0, l = 0, r = height.length - 1;
      while (l < r) {
          maxarea = Math.max(maxarea, Math.min(height[l], height[r]) * (r - l));
          if (height[l] < height[r])
              l++;
          else
              r--;
      }
      return maxarea;
  }
}



//我的解法,解法不正确
//我的思路:每次比较单侧,利用偏移量来比较,和官方解法对比后,感觉我的思路和暴力法更相似
public class BestWaterContainer {

	public static void main(String[] args) {
		//int[] heights= {10,9,8,7,6,5,4,3,2,1};//25
		//int[] heights= {1,8,6,2,5,4,8,3,7};//49
		//int[] heights= {1,1};
		//int[] heights= {1,2,1};
		int[] heights= {2,3,10,5,7,8,9};//36
		System.out.println(maxArea(heights));
	}

	public static int maxArea(int[] heights) {
		//初始化最大值
		int max = getAcreage(0, heights.length-1, heights);
		System.out.println("初始化的Max:"+ max);
		int i = 0;
		int j = heights.length - 1;
		int offsetLeft = 1;
		int offsetRight = 1;
		while ((i+offsetLeft)<=(j+offsetRight)&&(i+offsetLeft)<heights.length||(j-offsetRight)>= 0) {
			// 首先是0和lenght-1,然后让其分别与单侧相邻点比较,取大者,计算值,若值更大者留下
			if ((i+offsetLeft)<heights.length&&heights[i]< heights[i + offsetLeft]) {
				int area = getAcreage(i + offsetLeft, j, heights);
				if (area > max) {
					max = area;
					i = i + offsetLeft;
				}
				//System.out.println("循环中的max:i :"+max);
			}
			if ((j-offsetRight)>= 0&&heights[j] < heights[j - offsetRight]) {
				int area = getAcreage(i, j-offsetRight, heights);
				if (area > max) {
					max = area;
					j = j - offsetRight;
				}
				//System.out.println("循环中的max:j :"+max);
			}
			offsetLeft++;
			offsetRight++;
		}
		return max;

	}

	public static int getAcreage(int startPoint, int endPoint, int heights[]) {
		int height = Math.min(heights[startPoint], heights[endPoint]);
		int wide = endPoint - startPoint;
		return wide * height;
	}

}

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转载自blog.csdn.net/zcy_wxy/article/details/86223189