LeetCode解题报告--Container With Most Water

题目：最大的盛水容器

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.

原题链接地址：https://leetcode.com/problems/container-with-most-water/

分析：题意是给定长度为n整型数组height;分别构成(1,height[1]),(2,height[2]),….(n,height
)坐标点，找出两个点分别过这些点做垂直X轴的垂线，以及连接这两个点，与X轴构成长方形容器，求该容器的最大值！

通过两个变量leftHeiht,rightHeight分别指向数组height的最左端和最右端,这样宽度最大，当缩小宽度，则需要寻找较大的高度。

做简单证明



Java 代码：（accepted）

public class ContainerWithMostWater {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int[] a = { 1, 4, 5, 6, 3, 8, 9, 12, 45 };
System.out.println("Max Area is : " + maxArea(a));
int[] a1 = { 12,45,32,78,11,23,14 };
System.out.println("Max Area is : " + maxArea(a1));
int[] a2 = { 90,34,45,21,43,32};
System.out.println("Max Area is : " + maxArea(a2));
}

public static int maxArea(int[] height) {

int maxArea = 0; // The max area
int leftHight = 0;
int rigtHight = height.length - 1;

while (leftHight < rigtHight) {
//Calculate the max area
maxArea = Math.max(maxArea, (rigtHight - leftHight)
* Math.min(height[leftHight], height[rigtHight]));
//Because of cast effect, hence choose shortest height

if (height[leftHight] > height[rigtHight])
rigtHight--;
else
leftHight++;
//System.out.println("Left: " + height[leftHight] + " " + leftHight + " Right: " + height[rigtHight] + " " + rigtHight);
//System.out.println("Max Area: " + maxArea);
}
return maxArea;
}

}

测试结果：

Max Area is : 30
Max Area is : 92
Max Area is : 172

相关代码放在个人github:https://github.com/gannyee/LeetCode/tree/master/src