题目描述
Given an array nums of n integers where n > 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].
Example:
Input: [1,2,3,4]
Output: [24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
Qiang的思路
class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
res = [1 for _ in range(len(nums))]
for i in range(1, len(nums)):
res[i] = res[i-1] * nums[i-1]
tmp = 1
for i in range(len(nums)-2, -1, -1):
tmp = tmp * nums[i+1]
res[i] = tmp * res[i]
return res
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
vector<int> res(nums.size());
res[0]=1;
for(int i=1; i<nums.size(); i++)
res[i]=res[i-1]*nums[i-1];
int tmp=1;
for(int i=nums.size()-2; i>=0; i--){
tmp*=nums[i+1];
res[i]*=tmp;
}
return res;
}
};
网友评论