LeetCode--Array--Container With Most Water (Medium)

11. Container With Most Water (Medium)

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

Note: You may not slant the container and n is at least 2.

Example:

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

    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.

solution

The first solution: violent solution, but also the most likely to think of the solution

class Solution {
    public int maxArea(int[] height) {
        int max = 0;
        for (int i = 0; i < height.length; i++)
        {
            for (int j = i + 1; j <height.length; j++)
            {
                int s = 0;
                if (height[i] > height[j])
                    s = height[j] * (j - i);
                else
                    s = height[i] * (j - i);    
                max = s > max ? s : max;   
            }
        }
        return max;
    }
}

The second solution: greedy algorithm

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

reference
https://leetcode.com/articles/container-with-most-water/

to sum up

The first algorithm is very easy to think of violence, but the time complexity is O (n ^ 2); second using a greedy algorithm, time complexity is O (n).

Notes
1. greedy algorithm can be obtained on the basis of violence algorithms;