演算法參考：https://mp.weixin.qq.com/s?__biz=MzAxODQxMDM0Mw==&mid=2247484482&idx=1&sn=9503dae2ec50bc8aa2ba96af11ea3311&source=41#wechat_redirect

    public static int q() {
        int sumWater = 0;
        int[] a = {0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1};
        for (int i = 1; i < a.length - 1; i++) {
            
 
//左側最大值
            int highLeftMax = 0;
            //右側最大值
            int highRightMax = 0;
            int left = i - 1;
            int right = i + 1;
            //獲得左側最大值
            while (left >= 0) {
                highLeftMax = Math.max(highLeftMax, a[left]);
                left--;
            }
            
 
//獲取右側最大值
            while (right < a.length) {
                highRightMax = Math.max(highRightMax, a[right]);
                right++;
            }
            //同時滿足左側最大值大於當前值，並且右側最大值大於當前值，才可以接雨水
            if (highLeftMax > a[i] && highRightMax > a[i]) {
                int temp = Math.min(highLeftMax, highRightMax) - a[i];
                sumWater 
 
+= temp;
                System.out.println("i=" + i + ";值=" + a[i] + ";highLeftMax=" + highLeftMax + ";highRightMax=" + highRightMax + ";sumWater=" + temp);
            }
        }
        System.out.println(sumWater);
        return sumWater;
    }