11. Container With Most Water

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.

有个高度数组，就相当于隔板的高度，求数组中任意两隔板间盛水的最大量。隔板间的距离与较低隔板的高度乘积即为盛水的容量。

public class Solution {
    public int maxArea(int[] height) {
        int i = 0, j = height.length - 1, temp = 0;
        int maxArea = 0;
        if (height[i] < height[j]) {
            maxArea = (j - i) * height[i];
        } else {
            maxArea = (j - i) * height[j];
        }
        while (i < j) {
            if (height[i] < height[j]) {
                temp = (j - i) * height[i];
                i++;
            } else {
                temp = (j - i) * height[j];
                j--;
            }
            if (temp > maxArea) {
                    maxArea = temp;
            }
        }

        return maxArea;
    }
}