[LeetCode] 11. Container With Most Water

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.

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

盛最多水的容器。

给你 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0) 。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。

说明：你不能倾斜容器。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/container-with-most-water
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

存储水量 = 宽度 * 高度 = (right - left) * Math.min(height[left], height[right])

时间O(n)

空间O(1)

Java实现

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

JavaScript实现

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

LeetCode 题目总结