<p>

238. Product of Array Except Self

Total Accepted: 53516
Total Submissions: 121209
Difficulty: Medium

Given an array of n integers where n > 1,nums
, return an arrayoutput
such thatoutput[i]
is equal to the product of all the elements ofnums
exceptnums[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.)
</p>
<code>
public class ProductofArrayExceptSelf238 {

    public int[] productExceptSelf(int[] nums) {
        
        int[] res = new int[nums.length];
        int[] valR = new int[res.length];
        valR[valR.length-1]=1;
        for(int i=valR.length-2;i>=0;--i){//求得该位置的所有右边的数的乘积
            valR[i]=nums[i+1]*valR[i+1];
            System.out.print(" valR["+i+"]="+valR[i]);
        }
        System.out.println();
        res[0]=1;
        for(int i=1;i<res.length;i++){//求得所有左边，并求总乘积
            res[i]=res[i-1]*nums[i-1];
            
            System.out.print("--res["+(i-1)+"]="+res[i-1]);
            res[i-1] = res[i-1]*valR[i-1];
            System.out.print("++res["+(i-1)+"]="+res[i-1]);
        }
        return res;  
    }  

public static void main(String[] args) {
    // TODO Auto-generated method stub
    ProductofArrayExceptSelf238 pae = new ProductofArrayExceptSelf238();
    int[] nums ={1,2,3,4};
    
    
    int[] result = pae.productExceptSelf(nums);
    System.out.println();
    for(int i : result){
        System.out.print(i+" ");
    }

}

}
</code>