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 and n is at least 2.

The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

利用two pointers的方法，一開始指向左右邊界，利用比較矮的隔板高度計算當前能儲存的水量，然後比較矮的隔板往中間動一單位繼續掃描，直到左隔板和右隔板重合。

int maxArea(vector<int>& height) { // time: O(n); space: O(1)
    int res = 0, n = height.size(), l = 0, r = n - 1;
    while (l < r) {
        res = max(res, min(height[l], height[r]) * (r - l));
        if (height[l] <= height[r]) {
            ++l;
        } else {
            --r;
        }
    }
    return res;
}