【leetcode】11.Container With Most Water（c语言）

最新推荐文章于 2020-08-18 15:30:59 发布

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本文探讨了如何从一组垂直线中找出两线与x轴构成的最大面积容器。通过对比穷举法与贪心算法，详细阐述了每种方法的实现步骤及代码示例，最终得出贪心算法在效率上的优势。

Description：

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.

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Example1:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

[ 题目链接 ]

解法一：穷举

提交代码：

int maxArea(int* height, int heightSize)
{
	int i,j,shortline;
	int maxArea=0;

	for (i = 0; i <heightSize;i++)
		for (j = i + 1; j < heightSize; j++)
		{
			if (height[i] < height[j])	shortline = height[i];
			else shortline = height[j];
			if ((j - i)*shortline > maxArea)
				maxArea = (j - i)*shortline;
		}
	return maxArea;
}

运行结果：
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解法二：贪心算法
提交代码：

int maxArea(int* height, int heightSize)
{
	int left = 0, right = heightSize - 1;
	int shortline, currentArea, maxArea = 0;

	while (left != right)
	{
		if (height[left] > height[right])
		{
			shortline = height[right];
			currentArea = (right - left)*shortline;
			if (currentArea > maxArea)
				maxArea = currentArea;
			right--;
		}
		else
		{
			shortline = height[left];
			currentArea = (right - left)*shortline;
			if (currentArea > maxArea)
				maxArea = currentArea;
			left++;
		}
	}
	return maxArea;
}