The sword refers to offer (51) to build a product array

Topic description

Given an array A[0,1,...,n-1], please construct an array B[0,1,...,n-1], where element B[i]=A[ 0]*A[1]*...*A[i-1]*A[i+1]*...*A[n-1]. Division cannot be used.

 

topic analysis

There are two solutions to this problem, the second is an optimization based on the first one.

The first solution: B[i]=A[0]*A[1]*...*A[i-1]*A[i+1]*...*A[n-1]=C [i-1]*D[i-1], that is, we require C[i] and D[i], which is equivalent to sacrificing space complexity for time complexity.

The second solution: In the first solution, we found that an additional array is required, which is obviously unnecessary, because the calculation of the B array comes from A, and no redundant C and D arrays are needed, so what is better? way?

We can directly use the B array and implement it with the help of the intermediate variable tmp, see the code for details.

 

 

code

The first solution:

function multiply(array) {
    let C = [], D = [], n = array.length;
    C[0] = array[0];
    for (let i = 1; i < n; i++) {
        C[i] = array[i] * C[i - 1];
    }
    D[n-1] = array[n-1];
    for (let i = n - 2; i >= 0; i--) {
        D[i] = array[i] * D[i + 1];
    }
    let B = [];
    B[0]=D[1];
    B[n-1]=C[n-2];
    for (let i = 1; i < n - 1; i++) {
        B[i] = C[i-1] * D[i+1];
    }
    return B;
}

The second method:

function multiply(array) {
    let B = [], len = array.length;
    B[ 0]=1 ;
      // Calculate the product of the first i - 1 elements 
    for (let i = 1; i < len; i++ ) {
        B[i] = array[i-1] * B[i - 1];
    }
    let tmp =1 ;
      // calculate the product of the last N - i elements and concatenate 
    for (let i = len-2; i >=0; i-- ) {
        tmp*=array[i+1];
        B[i]*=tmp;
    }
    return B;
}