238. Product of Array Except Self

LeetCode Link
下面的 Solution 可以 optimize to O(1) space:

• leftToRight as res
• rightToLeft store with a temp variable

public int[] productExceptSelf(int[] nums) {
    int len = nums.length;
    int[] leftToRight = new int[len + 1];
    int[] rightToLeft = new int[len + 1];
    leftToRight[0] = 1;
    rightToLeft[len] = 1;

    for (int i = 0; i < len; i++) {
        leftToRight[i + 1] = leftToRight[i] * nums[i];
        rightToLeft[len - i - 1] = rightToLeft[len - i] * nums[len - i - 1];
    }

    int[] res = new int[len];
    for (int i = 0; i < len; i++) {
        res[i] = leftToRight[i] * rightToLeft[i + 1];
    }
    return res;
}