11. Container With Most Water

two pointers O(n)
这道题和maximum square不一样
思路和trapping most water类似，都是维护左右两个指针从外往里找短板，当前最大面积更新后，短板必须要更新成更长的板

class Solution:
    def maxArea(self, height: List[int]) -> int:
        l, r, res = 0, len(height) - 1, 0
        while l < r:
            mn = min(height[l], height[r])
            res = max(res, mn * (r - l))
            while l < r and height[l] <= mn:
                l += 1
            while l < r and height[r] <= mn:
                r -= 1
        return res

class Solution {
public:
    int maxArea(vector<int>& height) {
        int n = height.size();
        int res = 0;
        int l = 0, r = n - 1;
        while (l < r) {
            int mn = min(height[l], height[r]);
            res = max(res, mn * (r - l));
            while (l < r && height[l] <= mn) {
                ++l;
            }
            while (l < r && height[r] <= mn) {
                --r;
            }
        }
        return res;
    }
};