11. Container With Most Water

用two pointers夹逼的方法，刚开始把left和right设置在数组的两端，如果left处比right处矮，那么说明left处是瓶颈，否则就是right处是。

同时维护一个全局的maxArea

    public int maxArea(int[] height) {
        if(height == null || height.length < 2) {
            return 0;
        }
        int maxArea = 0;
        int len = height.length;
        int right = len - 1;
        int left = 0;
        while(left < right) {
            int area = (right - left) * Math.min(height[left], height[right]);
            maxArea = Math.max(maxArea, area);
            if(height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        return maxArea;
    }