#leetcode#Product of Array Except Self

Product of Array Except Self

Total Accepted: 442 Total
Submissions: 1138My Submissions

Question
 Solution

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.)

sdf

前几天刚看facebook的面经出现这题， leetocde就加上了， 不能用除法， 则维护当前元素左边全部元素的乘积以及右边全部元素的乘积， 相乘得到 product of array except self !

Because we cannot use division, so assume we have two integer arrays with the same length of nums, int[]
leftProd = new int[nums.length]; int[] rightProd = new int[nums.length], we store the product of all the left elements in leftProd and
the product of all the right elements in rightProd,
then the product of leftProd[i] and rightProd[i] will
be the value we want to put into the result. take the example of num[] = {2, 4, 3, 6},
thenleftProd will
be {1, 2, 8, 24} ,
and rightProd will be {72, 18, 6, 1}.

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        if(nums == null)
            return null;
        int[] res = new int[nums.length];
        for(int i = 0; i < nums.length; i++){
            if(i == 0)
                res[i] = 1;
            else
                res[i] = res[i - 1] * nums[i - 1];
        }
        int prod = 1;
        for(int i = nums.length - 1; i >= 0; i--){
            res[i] = res[i] * prod;
            prod *= nums[i];
        }
        return res;
    }
}