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;
}
}