README_EN.md

11. Container With Most Water

中文文档

Description

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 the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

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

Example 3:

Input: height = [4,3,2,1,4]
Output: 16

Example 4:

Input: height = [1,2,1]
Output: 2

 

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
        res = 0
        while i < j:
            t = (j - i) * min(height[i], height[j])
            res = max(res, t)
            if height[i] < height[j]:
                i += 1
            else:
                j -= 1
        return res

Java

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

C++

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

Go

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

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

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

JavaScript

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

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