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 count as extra space for the purpose of space complexity analysis.)
思路
从前向后累乘一遍，从后向前累乘一遍，一开始是用的两个数组，高分答案中遍历一遍即可。
代码
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int len=nums.size();
vector<int> res(len, 1);
for(int i=1; i<len; i++){
res[i] = nums[i-1]*res[i-1];
res[len-i-1] = nums[len-i]*res[len-i];
}
return res;
}
};
一遍过
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n=nums.size();
int fromBegin=1;
int fromLast=1;
vector<int> res(n,1);
for(int i=0;i<n;i++){
res[i]*=fromBegin;
fromBegin*=nums[i];
res[n-1-i]*=fromLast;
fromLast*=nums[n-1-i];
}
return res;
}
};
