Container With Most Water
problem description and solution for the LeetCode problem "Container With Most Water"
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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])
.
您将获得一个长度 n
的整数数组 height
。绘制了 n
垂直线,使得 ith
线的两个端点是 (i, 0)
和 (i, height[i])
。
Find two lines that together with the x-axis form a container, such that the container contains the most water.
找到两条与 x 轴一起形成容器的线,这样容器包含的水最多。
Return the maximum amount of water a container can store.
返回容器可以储存的最大水量。
Notice that you may not slant the container.
请注意,您不能倾斜容器。
Example 1: 示例 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: 示例 2:
Input: height = [1,1] Output: 1
Constraints: 约束:
n == height.length
2 <= n <= 105
0 <= height[i] <= 104
package org.leetcode;
public class ContainerWithMostWater {
public int maxArea(int[] height) {
int area = 0;
int left = 0;
int right = height.length - 1;
while(left < right) {
int currentHeight = Math.min(height[left], height[right]);
int currentWidth = right - left;
int currentArea = currentHeight * currentWidth;
if(currentArea > area) {
area = currentArea;
}
if(height[left] < height[right]) {
left++;
} else {
right--;
}
}
return area;
}
public static void main(String[] args) {
ContainerWithMostWater containerWithMostWater = new ContainerWithMostWater();
int[] height1 = {1,8,6,2,5,4,8,3,7};
int result1 = containerWithMostWater.maxArea(height1);
int expected1 = 49;
if(result1 == expected1) {
System.out.println("Test 1 passed");
} else {
System.out.println("Test 1 failed expected: " + expected1 + " actual: " + result1);
}
}
}