Source:https://leetcode.com/problems/container-with-most-water/
Time:2020/12/20
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 <= 3 * 104
0 <= height[i] <= 3 * 104
Solution
Approach1: Bruce force
class Solution {
public int maxArea(int[] height) {
int res = -1;
for(int i = 0; i < height.length; i++)
for(int j = i; j < height.length; j++)
{
int h = Math.min(height[i], height[j]);
int a = j - i;
res = Math.max(res, a * h);
}
return res;
}
}
Approach 2: Two Pointer Approach
But moving the shorter line’s pointer could turn out to be beneficial, as per the same argument, despite the reduction in the width.
class Solution {
public int maxArea(int[] height) {
int res = -1;
int l = 0, r = height.length - 1;
while(l < r)
{
res = Math.max((r- l) * Math.min(height[l], height[r]), res);
if(height[l] < height[r])
l++;
else
r--;
}
return res;
}
}