Problem:
Analysis:
Using two round processing: forward and backward. See figure for detail:
//////////////////////////////////////////
//code 60ms
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
vector<int> tmp(nums.size(), 1);
int n=nums.size();
//forward
for(int i=0;i<n-1;i++){
tmp[i+1]=tmp[i]*nums[i];
}
//backward
int m=1;
for (int j=n-1;j>-1;j--){
tmp[j]=m*tmp[j];
m=m*nums[j];
}
return tmp;
}
};
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
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.)
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Analysis:
Using two round processing: forward and backward. See figure for detail:
//////////////////////////////////////////
//code 60ms
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
vector<int> tmp(nums.size(), 1);
int n=nums.size();
//forward
for(int i=0;i<n-1;i++){
tmp[i+1]=tmp[i]*nums[i];
}
//backward
int m=1;
for (int j=n-1;j>-1;j--){
tmp[j]=m*tmp[j];
m=m*nums[j];
}
return tmp;
}
};