238 title: in addition to its own product array

One. Problem Description

A given length n integer array nums, wherein n> 1, the output returns an array output, wherein the output [i] is equal to the product of the remaining element other than nums [i] nums in.

Example:

Input: [1,2,3,4]

Output: [24,12,8,6]

Note: Do not use a divider, and completed this title in (n) time complexity O.

Advanced:

You can complete this topic yet in constant space complexity? (For the purpose of the space complexity analysis, output array is not considered extra space)

two. Problem-solving ideas

The title ideas: The title dual pointer way to solve, since the subject of the request, please do not use a divider, and in (O n- complete problem within) time complexity, we should take into account the traverse often several times, using the method of the left and right hands will be solving.

Step a: Create two arrays of length n is initialized to the fir and sec 1.

Step Two: Create about pointers, and the second number is traversed from the penultimate number,

for [j] = for [j-1] * nums [j-1];

           sec[m]=sec[m+1]*nums[m+1];

Step 3: the array in the array is multiplied by two, the final array output.

three. Results of the

When execution: 1 ms, defeated 100.00% of users in all java submission

Memory consumption: 43.2 MB, defeated 96.20% of all users in java submission

four. Java code

class Solution {
    public int[] productExceptSelf(int[] nums) {
         int []fir=new int[nums.length];
        int []sec=new int[nums.length];
        for(int i=0;i<nums.length;i++) {
            for [i] = 1 ;
            sec[i]=1;
        }
        for(int j=1,m=nums.length-2;j<nums.length&&m>=0;j++,m--) {
            for [j] = for [j-1] * nums [j-1 ];
            sec[m]=sec[m+1]*nums[m+1];
        }
        for(int k=0;k<nums.length;k++) {
            for [k] = for [k] * sec [k];
        }
        return fir;
    }
}