LeetCode | Container With Most Water

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博客围绕LeetCode 11题“Container With Most Water”展开，给定n个非负整数代表坐标点，要找出两条线与x轴构成盛水最多的容器。给出两种Java解法，一是用两个for循环计算面积，时间复杂度O(n²)；二是用头尾指针，小指针往中间缩，时间复杂度O(n)。

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11. Container With Most Water

Description

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 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.

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

Solution: (Java)

class Solution {
    public int maxArea(int[] height) {
        int most = 0, container;
        for (int i = 0; i < height.length - 1; i++) {
            for (int j = i + 1; j < height.length; j++) {
                container = (j-i)*(Math.min(height[j], height[i]));
                if (container > most) {
                    most = container;
                }
            }
        }
        return most;
    }
}

思路：
这里直接用两个 for 循环，逐次计算区域面积，保存最大的即可，时间复杂度为 O(n²)，accept 之后看了其他的思路，即下面方法2的思路：用头尾两个指针，对应高度较小的指针往中间缩减（只由较小的长度决定面积），依次计算区域面积，时间复杂度为 O(n).

方法2:

public int maxArea2(int[] height) {
    int most = 0, left = 0, right = height.length - 1;
    while (left < right) {
        most = Math.max(most, (right-left) * (Math.min(height[left], height[right])));
        if (height[left] < height[right])
            left++;
        else
            right--;
    }
    return most;
}