[Offer] to prove safety of building product array

Title Description

Given an array A [0,1, ..., n- 1], please construct an array B [0,1, ..., n- 1], where B is the element B [i] = A [ 0] * A [1] * ... * A [i-1] * A [i + 1] * ... * A [n-1]. You can not use the division .

Analysis: Note that the subject of the request can not be used division!

We found that if action is taken to violence, then there will be a lot of repetitive operations, we can use an array of records to reduce repetitive operation

Using a two-dimensional array to record the product and dp, dp representatives [i] [j] to the index i of the product of the index j , after the calculation may be recorded, so that when later need no longer required to calculate, it can be directly used, this is the so-called memorandum of law

First of dp array initialization

dp[0][0]=a[0]

dp[0][i]=dp[0][i-1]*a[i],i>=1,i<n

dp[n-1][n-1]=a[n-1]

dp[i][n-1]=dp[i+1][n-1]*a[i]

State transition formula:

When i == 0, b [i] = dp [1] [n-1], a separate header processing unit

When when i == n-1, b [i] = dp [0] [i-1], the tail alone treatment

i>0&&i<n-1时，b[i]=dp[0][i-1]*dp[i+1][n-1]

Time complexity: O (N)

Space complexity: O (N * N)

class Solution 
{ 
public : 
    Vector < int > Multiply ( const Vector < int > & A) 
    { 
        int n-= a.size ();
         int [n-] DP [n-]; // DP [i] [J]: The index i the product of the element index j 
        Memset (DP, 0 , the sizeof ( 0 )); 
        Vector < int > B (n-, 0 ); 
        
        // initialize 
        DP [ 0 ] [ 0 ] = a [ 0 ];
         for ( int= I . 1 ; I <n-; I ++ ) 
        { 
            DP [ 0 ] [I] DP = [ 0 ] [I- . 1 ] * A [I]; 
        } 
        DP [n- - . 1 ] [N- . 1 ] = A [n- - . 1 ];
         for ( int I = N- 2 ; I> = 0 ; i-- ) 
        { 
            DP [I] [n- - . 1 ] DP = [I + . 1 ] [N- . 1 ] * A [I]; 
        } 
        
        // need to handle a separate header and trailer 
        for ( int I = 0; i<n; i++)
        {
            if(i==0)
            {
                b[i]=dp[1][n-1];
            }
            else if(i==n-1)
            {
                b[i]=dp[0][i-1];
            }
            else
            {
                b[i]=dp[0][i-1]*dp[i+1][n-1];
            }
        }
        return b;
    }
};