class Solution {
    public int[] constructArr(int[] a) {
        if (a.length == 0) return new int[]{};
        if (a.length < 2) {
            return new int[]{0};
        }
        int n = a.length;

        int[] mulLeftToRight = new int[n];
        int[] mulRightToleft = new int[n];

        mulLeftToRight[0] = a[0];
        mulRightToleft[n - 1] = a[n - 1];

        for (int i = 1; i < n - 1; i++) {                           //从左往右的乘积
            mulLeftToRight[i] = mulLeftToRight[i - 1] * a[i];
        }

        for (int i = n - 2; i > 0; i--) {                           //从右往左的乘积
            mulRightToleft[i] = mulRightToleft[i + 1] * a[i];
        }


        //处理结果的边界条件
        a[0] = mulRightToleft[1];
        a[n - 1] = mulLeftToRight[n - 2];

        //非边界区域的值可由左边的值乘以右边的值得到
        for (int i = 1; i < n - 1; i++) {
            a[i] = mulLeftToRight[i-1]*mulRightToleft[i+1];
        }

        return a;
    }
}