Container With Most Watert

题意：有n个相当于水桶的隔板，求使用两个隔板和底部x轴所能组成的水桶可以装载的最大容量。

第一种方法：两层遍历，第一层从0遍历到n-2，第二层从i+1遍历到n-1，每次求得（j-i）*（Math.min(height【i】，height【j】)）与max比较，使max总是保存最大值。

缺点：需要n*n时间

第二种方法：线性时间，两个指针分别从0、n-1开始向中间靠拢，直到左指针不小于右指针。

靠拢规则：左高渡小于右高度，则左指针自加；否则右指针自加。因为最大值的产生肯定在同等宽度条件下的高度最大值。

时间：线性，代码如下：

/** * https://leetcode.com/problems/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. */package leetcode;public class MaxArea {public static void main(String[] args) {}public static int maxArea(int[] height) {        if (height==null || height.length<=0) {return 0;}                int left = 0, right = height.length - 1;        int max = 0;                while (left < right) {        int temp = (right-left)*Math.min(height[left], height[right]);        if (temp > max) {max = temp;}        if (height[left] < height[right]) {left++;} else {right--;}        }                return max;    }}