每周算法:最大容器

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.

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

来源: leetcode 11 Container with most water

Solution

Approach 1 暴力解法

暴力解法是最容易想到的解法了,然而暴力解法肯定不是本题的意图。

int maxArea(vector<int>& height) {
    int ans = 0;

    int len = height.size();

    for (int i=0; i < len-1; ++i) {
	for (int j=i+1; j < len; ++j) {
	    int size = (j-i) * min(height[i], height[j]);
	    ans = ans > size ? ans : size;
	}
    }

    return ans;
}

使用这种解法只要把变量变化的范围想清楚就行了,没有什么特别的难点。
暴力解法的时间复杂度为 O(n2),空间复杂度为 O(1)。

Approach 2

我在leetcode上看到这样一种解法,这种解法从逻辑和形式上看上不复杂,但是我的疑问在是否有严密的数学推导和论证。

int maxArea(vector<int>& height) {

    int ans = 0;
    int left = 0;
    int right = height.size()-1;

    while (left < right) {
	int w = right - left;
	int h = min(height[left], height[right]);

	ans = max(w*h, ans);

	if (height[left] < height[right]) {
	    ++left;
	}
	else {
	    --right;
	}
    }

    return ans;
}

这个算法的主要思想是最大面积由两端边界较短的板决定;所以在移动边界时要将长板保留,移动短板。该算法的时间复杂度为 O(n),空间复杂度为O(1)。

在寻求该算法的数学推导时,我看到了 @nan15 给出的另外一种解释:

We starts with the widest container, l = 0 and r = n - 1 . Let's say the left one is shorter: h[l] < h[r] . Then, this is already the largest container the left one can possibly form. There's no need to consider it again. Therefore, we just throw it away and start again with l = 1 and r = n - 1 .

这个论证我认为是更加直观的:从最宽的容器开始,这时左边坐标 l = 0 ,右边坐标 r = n - 1 。假设左边更短,也就是 h[l] < h[r] 。那么,对于左边的板子,这是已经是它能够形成的最大的面积了,也就不需要再考虑它了。因此,将左边的板子丢掉,使用 l = 1r = n - 1 重新开始。

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