[LeetCode] 11. Container With Most Water

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

Example 1:

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

Constraints:
n == height.length
2 <= n <= 105
0 <= height[i] <= 104

盛最多水的容器。

给定一个长度为 n 的整数数组 height 。有 n 条垂线，第 i 条线的两个端点是 (i, 0) 和 (i, height[i]) 。
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。
返回容器可以储存的最大水量。
说明：你不能倾斜容器。
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/container-with-most-water
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思路

双指针，两边往中间逼近。逼近的条件是哪边的高度比较矮，就移动哪个指针。
存储水量 = 宽度 * 高度 = (right - left) * Math.min(height[left], height[right])

复杂度

时间O(n)
空间O(1)

代码

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;
};