leetcode.11---------Container With Most Water

寻找最大水容器算法

最新推荐文章于 2022-03-29 21:43:55 发布

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本文介绍了一种寻找最大水容器的算法问题，该问题要求在给定一系列高度的垂直线中找到两根线，这两根线与x轴形成的矩形面积最大。文中提供了两种解决方案，一种是暴力穷举的方法，另一种则是使用双指针技巧逐步逼近最优解。

题目：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.

题目的意思大概就是：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|

方法一：暴力穷举方法

class Solution {
public:
	int maxArea(vector<int> &height)
	{
		int n = height.size();
		int max = 0;
		for (int i = 0; i < n; i++)
		{
			for (int j = i + 1; j < n; j++)
			{
				int temp = (j - i)*(height[i]>height[j] ? height[j] : height[i]);
				if (temp > max)
				{
					max = temp;
					continue;
				}
			}
		}
		return max;
	}
};

方法二：

The basic idea is simple: use two pointers to indicate the left edge and right edge of the water container. Every time move the lower edge towards another edge with one step. At last, we can find the max water container.

class Solution {
public:
	int maxArea(vector<int> &height) {
		int start = 0;
		int end = height.size() - 1;
		int result = INT_MIN;
		while (start < end)
		{
			int area = min(height[end], height[start]) * (end - start);
			result = max(result, area);
			if (height[start] <= height[end])
			{
				start++;
			}
			else
			{
				end--;
			}
		}
		return result;
	}
};

改进：

int maxArea(vector<int> &height) {
    int left = 0, right = height.size() - 1;
    int maxWater = 0;
    while(left < right){
        maxWater = max(maxWater, (right - left) * min(height[left], height[right]));
        height[left] < height[right] ? left++ : right--;
    }
    return maxWater;
}