Container With Most Water

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

Given [1,3,2], the max area of the container is 2.

Analysis:

Begin with the first point and last point, calculate the container area. Move the first point forward or last point backward if it is smaller.

 1 public class Solution {
 2   public int maxArea(int[] height) {
 3     // corner case
 4     if (height == null || height.length == 1) {
 5       return 0;
 6     }
 7 
 8     int pStart = 0, pEnd = height.length - 1, maxArea = 0;
 9     while (pStart < pEnd) {
10       int temp = (pEnd - pStart) * Math.min(height[pStart], height[pEnd]);
11       maxArea = Math.max(maxArea, temp);
12 
13       // if the lenght of the line pointed by pStart is less than the length of the
14       // line pointed by pEnd we should increate pStart by 1, otherwise, decrease pEnd by 1.
15       if (height[pStart] < height[pEnd]) {
16         pStart++;
17       } else {
18         pEnd--;
19       }
20     }
21     return maxArea;
22   }
23 }

 

转载于:https://www.cnblogs.com/beiyeqingteng/p/5669071.html