LeetCode-array-Container With Mo

首先看题目：
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.

意思是，有一个非负数的数组，数组中每个元素就是一条垂直的线，这条线的两端坐标是（i,ai）,( i,0)，i是数组元素的下标，
选取两条组成面积最大的，计算出值，注意，矩形不能倾斜，，，

1499077569(1).jpg

如上图。
从java代码解决，直接上代码

/**
 * Created by Wangjianxin on 2017/7/3 0003.
 */
public class ContainerWithMostWater {

    public int maxArea(int[] height) {
        int oldmax = 0;  
        int newmax = 0;
        int len = height.length;

        int left = 0;
        int right = len - 1;

        while (left < right){ 
            //计算面积，看上图可以知道，需要两条线中短线的高，和两线之间的距离宽
            if(height[left] > height[right]){ 
                oldmax = height[right] * (right - left);
            }else {
                oldmax = height[left] * (right - left);
            }

            if(newmax < oldmax){
                newmax = oldmax; //得到最大值
            }
            //这里用于从两头往中间一直计算，只要两条线中壹条短一些， 就应该在那条短的线继续向前或者向后的其他线在计算
            if(height[left] > height[right]){
                right --;
            }else {
                left ++;
            }

        }
        return newmax;
    }
}