Leetcode 238 Product of Array Except Self

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:
Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

public class S238 {
    public int[] productExceptSelf(int[] nums) {
        //左右两个数组分别表示nums[i]左右两边的乘积，那么rets[i]=left[i]*right[i]
        //判断是否为零反而时间变长，本来想法是判断到零就可以结束循环（零以后的数除自己以外的乘积都为零），减少复杂度
/*        int []left = new int[nums.length];
        left[0] = 1;
        int []right = new int[nums.length];
        right[nums.length-1] = 1;
        int []rets = new int[nums.length];
        for(int i = 1;i<nums.length;i++){
            left[i] = left[i-1]*nums[i-1]; 
            right[nums.length-i-1] = right[nums.length-i]*nums[nums.length-i];
            if(i>nums.length/2-1){
                rets[i] = left[i]*right[i];
            }
            if(nums[i] == 0){
                break;
            }
        }
        for(int i = 0;i<nums.length/2;i++){
            rets[i] = left[i]*right[i];
            if(nums[i] == 0){
                break;
            }
        }
        return rets;*/
        //减小空间复杂度的方法,同时也降低了运行时间，是否判断零没有影响
        int []rets = new int[nums.length];
        rets[nums.length-1] = 1;
        //先求出每个数右侧乘积
        for(int i = nums.length-2;i>=0;i--){
            rets[i] = rets[i+1]*nums[i+1];
            if(nums[i] == 0){
                break;
            }
        }
        //再从左往右，求左侧积的时候用rets保存结果
        int left = 1;
        for(int i = 0;i<nums.length;i++){
            rets[i] *= left;
            left *= nums[i];
        }
        return rets;
    }
}