11. Container With Most Wate(盛最多水的容器)

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

给定 n 个非负整数 a1，a2，...，an，每个数代表坐标中的一个点 (i, ai) 。在坐标内画 n 条垂直线，垂直线 i 的两个端点分别为 (i, ai) 和 (i, 0)。找出其中的两条线，使得它们与 x 轴共同构成的容器可以容纳最多的水。

说明：你不能倾斜容器，且 n 的值至少为 2。

图中垂直线代表输入数组 [1,8,6,2,5,4,8,3,7]。在此情况下，容器能够容纳水（表示为蓝色部分）的最大值为 49。

示例:

输入: [1,8,6,2,5,4,8,3,7]
输出: 49

class Solution {
    public int maxArea(int[] height) {
        int left = 0, right = height.length - 1;
        int max = 0;
        while (left < right) {
            int h = Math.min(height[left], height[right]); // 最短木板决定体积
            int l = right - left; // 长度
            int v = h * l; // 符合题意的平面体积
            max = Math.max(max, v);
            if (height[left] < height[right]) {
                ++left;
            } else {
                --right;
            }
        }
        return max;
    }
}

Debug code in playground:

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

public class MainClass {
    public static int[] stringToIntegerArray(String input) {
        input = input.trim();
        input = input.substring(1, input.length() - 1);
        if (input.length() == 0) {
          return new int[0];
        }
    
        String[] parts = input.split(",");
        int[] output = new int[parts.length];
        for(int index = 0; index < parts.length; index++) {
            String part = parts[index].trim();
            output[index] = Integer.parseInt(part);
        }
        return output;
    }
    
    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String line;
        while ((line = in.readLine()) != null) {
            int[] height = stringToIntegerArray(line);
            
            int ret = new Solution().maxArea(height);
            
            String out = String.valueOf(ret);
            
            System.out.print(out);
        }
    }
}

==========================Talk is cheap, show me the code========================