Leetcode 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.

Solution: 

Use two pointers. Set start pointer at index 0 and end at height.length - 1. The height is always the smaller one of start and end and the width is end - start. The volume is height * width. Then to find a larger volume while narrowing the width, we should
consider to increase the height after each move. So we should move the pointer with smaller value and try to update the volume until start equals to end.

Time complexity: O(n)

Space complexity: O(1)

<p>public class Solution {
public int maxArea(int[] height) {
int length = height.length;
int volume = Integer.MIN_VALUE;
if (height == null || length == 0) {
return 0;
}
// Two pointers
int i = 0;
int j = length - 1;
while (i < j) {
int wid = j - i;
int hei = Math.min(height[i], height[j]);
volume = Math.max(volume, wid * hei);
if (height[i] <= height[j]) {
i++;
} else {
j--;
}
}
return volume;
}
}
</p>