0011. Container With Most Water (M)

Container With Most Water (M)

题目

Given n non-negative integers \(a_1, a_2, ..., a_n\) , where each represents a point at coordinate \((i, a_i)\). n vertical lines are drawn such that the two endpoints of line i 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 and n is at least 2.

The above vertical lines are represented by array [1, 8, 6, 2, 5, 4, 8, 3, 7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

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

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题意

给出n个坐标点 \((i, a_i)\) ，每个坐标点与点 \((i, 0)\) 构成一条直线，任选两条直线，与x轴构成一个盛水的容器，求这个容器的最大容积。

思路

暴力法\(O(N^2)\)。

Two Pointers - \(O(N)\) : 设两指针分别指向数组的左右两端，计算容积并更新最大容积，移动高度较短的指针进行下次循环。证明：影响容器容积的因素有两个，即底边长度和较短边高度，对于高度较短的一方，它已经达到了能够构成的最大容积(因为底边长度已经是最大)，只有更换较短边才有可能抵消缩短底边长度带来的影响。
具体证明方法：过冰峰 - container-with-most-water(最大蓄水问题)

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代码实现

Java

class Solution {
    public int maxArea(int[] height) {
        int maxVolume = 0;
        int i = 0, j = height.length - 1;
        while (i < j) {
            int volume = (j - i) * Math.min(height[i], height[j]);
            maxVolume = Math.max(maxVolume, volume);
            if (height[i] < height[j]) {
                i++;
            } else {
                j--;
            }
        }
        return maxVolume;
    }
}

JavaScript

/**
 * @param {number[]} height
 * @return {number}
 */
var maxArea = function (height) {
  let left = 0, right = height.length - 1
  let max = 0
  while (left < right) {
    let h = Math.min(height[left], height[right])
    max = Math.max(max, (right - left) * h)
    if (h === height[left]) {
      left++ 
    } else {
      right--
    }
  }
  return max
}