Container With Most Water

最新推荐文章于 2024-08-28 08:50:09 发布

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本文介绍了一个经典的算法问题：给定一系列垂直线段，求解能够组成最大面积容器的两条线段。通过双指针技巧逐步缩小搜索范围，最终找到最优解。

Given n non-negative integers a₁, a₂, ...,a_n, where each represents a point at coordinate (i,a_i).n vertical lines are drawn such that the two endpoints of linei is at (i,a_i) 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.

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给定n个非负整数a1，a2，...，an，其中每个表示坐标（i，ai）处的点。绘制n条垂直线，使得线i的两个端点在（i，ai）和（i，0）。找到两条线，它们与x轴一起形成一个Container ,即：就是说，x轴上在1,2,...,n点上有许多垂直的线段，长度依次是a₁, a₂, ..., an。找出两条线段，使他们和x抽围成的面积最大。面积公式是 Min(a_i, a_j) X |j - i|

package leetcode;

/**
 * 
 * @author Mouse
 *
 */
public class Solution {
	public static int maxArea(int[] height) {
		int start = 0;
		int end = height.length-1;
		int max = 0;
		while (start < end) {
			int temp = (end - start)
					* (height[start] >height[end] ? height[end]
							: height[start]);
			if (temp > max) {
				max = temp;
			}
			if (height[start]<height[end]) {
				int k=start;
				while (k<end&&height[k]<=height[start]) k++;
				start=k;
			}else{
				int k=end;
				while (k>start&&height[k]<=height[end]) k--;
				end=k;
			}

		}
		return max;

	}

	public static void main(String[] args) {
		int[] height = { 2, 3, 4, 5, 18, 17,6};
		int max=maxArea(height);
		System.out.println(max);
	}

}