README_EN.md

11. Container With Most Water

中文文档

Description

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

Solutions

Python3

class Solution:
    def maxArea(self, height: List[int]) -> int:
        i, j = 0, len(height) - 1
        ans = 0
        while i < j:
            t = (j - i) * min(height[i], height[j])
            ans = max(ans, t)
            if height[i] < height[j]:
                i += 1
            else:
                j -= 1
        return ans

Java

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

C++

class Solution {
public:
    int maxArea(vector<int>& height) {
        int i = 0, j = height.size() - 1;
        int ans = 0;
        while (i < j) {
            int t = min(height[i], height[j]) * (j - i);
            ans = max(ans, t);
            if (height[i] < height[j]) {
                ++i;
            } else {
                --j;
            }
        }
        return ans;
    }
};

Go

func maxArea(height []int) (ans int) {
	i, j := 0, len(height)-1
	for i < j {
		t := min(height[i], height[j]) * (j - i)
		ans = max(ans, t)
		if height[i] < height[j] {
			i++
		} else {
			j--
		}
	}
	return
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

JavaScript

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

TypeScript

function maxArea(height: number[]): number {
    let i = 0;
    let j = height.length - 1;
    let ans = 0;
    while (i < j) {
        const t = Math.min(height[i], height[j]) * (j - i);
        ans = Math.max(ans, t);
        if (height[i] < height[j]) {
            ++i;
        } else {
            --j;
        }
    }
    return ans;
}

C#

public class Solution {
    public int MaxArea(int[] height) {
        int i = 0, j = height.Length - 1;
        int ans = 0;
        while (i < j) {
            int t = Math.Min(height[i], height[j]) * (j - i);
            ans = Math.Max(ans, t);
            if (height[i] < height[j]) {
                ++i;
            } else {
                --j;
            }
        }
        return ans;
    }
}

Rust

impl Solution {
    pub fn max_area(height: Vec<i32>) -> i32 {
        let mut i = 0;
        let mut j = height.len() - 1;
        let mut res = 0;
        while i < j {
            res = res.max(height[i].min(height[j]) * (j - i) as i32);
            if height[i] <= height[j] {
                i += 1;
            } else {
                j -= 1;
            }
        }
        res
    }
}

...