Algorithm Container With Most Water

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 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:

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

Submission

package com.cctoken.algorithm;

/**
 * @author chenchao
 */
public class ContainerWithMostWater {

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

  public static void main(String[] args) {

  }
}

Solution

计算任意两个边界能容纳最多的水，即是求最大面积，假设两个边界的横向索引位置分别为 i,j, 那么此时的面积为 Min(height[i],height[j])*(j-i)，我们分别从两次一次向中间递推，如果我们想获取更大的
面积，那么我们可以基于这样的一个认知，即放弃height比较低的一侧，向中间递增或者递减，那么不难得到上面的解答