Question 52: Building a product array

Topic:

Given an array a[0,1,..., n-1], build an array b[0,1,..., n-1], where the elements in B b[i]=a[0]*a[1]*...*a[i-1]*a[i+1]*...*a[n-1]. You cannot use division.

Ideas:

Method 1:

Direct multiplication n-1 number, get b[i];

Complexity of Time: O (n^2)

Method 2:

Constructs a forward product array c[i]=a[0]*a[1]*...*a[i-1], or c[i]=c[i-1]*a[i-1];

Build the back-to-product array d[i]=a[n-1]*a[n-2]* ... A[n-i+1], i.e. d[i]=d[i+1]*a[i+1];

by C[i],d[i] to seek b[i]:b[i]=c[i]*d[i]

Time complexity: O (N)

Code:

void Multiply (const vector<double>& array1,vector<double>& array2) {    int len1=array1.size ();    int len2=array2.size ();    if (len1==len2 && len2>1) {        array2[0]=1;        for (int i=1;i<len1;i++) {            array2[i]=array2[i-1]*array1[i-1];        }        Double tmp=1;        for (int i=len1-2;i>=0;i--) {            tmp*=array1[i+1];            array2[i]*=tmp;}}}    

Online test OJ:

Http://www.nowcoder.com/books/coding-interviews/94a4d381a68b47b7a8bed86f2975db46?rp=3

AC Code:

class solution {public:vector<int> Multiply (const vector<int>&        A) {int len=a.size ();        if (len<1) return vector<int> ();                Vector<int> B (len,1);        for (int i=1;i<len;i++) {b[i]*=b[i-1]*a[i-1];        } int tmp=1;            for (int i=len-2;i>=0;i--) {tmp*=a[i+1];        b[i]*=tmp;        } return B; }};

Class Solution {public:    vector<int> Multiply (const vector<int>& A) {    int len=a.size ();        if (len<1)            return vector<int> ();        Vector<int> B (len,1);        Vector<int> Front (len,1);        Vector<int> back (len,1);                for (int i=1;i<len;i++) {            front[i]=front[i-1]*a[i-1];        }                for (int i=len-2;i>=0;i--) {            back[i]=back[i+1]*a[i+1];        }                for (int i=0;i<len;i++)            b[i]=front[i]*back[i];                return B;        };

Question 52: Building a product array