Container With Most Water

,

Problem

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.

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

1
2

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

My Answer

  • height 배열을 순회 하면서 각 영역 별 최대 값을 구하고, 비교한다.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20

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

Fastest Answer

  • 좌, 우 끝단 부터 시작 해서 최대 값을 구한다.
  • Explain
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

class Solution {
    public int maxArea(int[] height) {
        int left=0;
        int right=height.length-1;
        int most=0;

        while (left < right) {
            int min = height[left] < height[right] ? height[left] : height[right]; 
            most = Math.max(most, min * (right - left)); 

            if (height[left] < height[right]) {
                int oldLeft = height[left]; 
                while (height[left] <= oldLeft && left < right) {
                    left++;
                }
            }else{
                int oldRight = height[right];
                while (height[right] <= oldRight && left < right) {
                    right--;
                }
            }
        }
        return most; 
    }
}

참고