【刷题-LeetCode】238. Product of Array Except Self
- Product of Array Except Self
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]
Constraint: It's guaranteed that the product of the elements of any prefix or suffix of the array (including the whole array) fits in a 32 bit integer.
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.)
解 left-right。计算第i个数左边和右边的乘积
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
vector<int>ans;
vector<int>left(nums.size(), 1), right(nums.size(), 1);
for(int i = 1; i < nums.size(); ++i){
left[i] = left[i-1] * nums[i-1];
}
for(int i = nums.size()-1; i > 0; --i){
right[i-1] = right[i]*nums[i];
}
for(int i = 0; i < nums.size(); ++i){
ans.push_back(left[i]*right[i]);
}
return ans;
}
};
优化:在从右往左扫描时,right数组中的数字只用了一次,因此用一个变量R代替即可
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
vector<int>ans(nums.size(), 1);
for(int i = 1; i < nums.size(); ++i){
ans[i] = ans[i-1] * nums[i-1];
}
int R = 1;
for(int i = nums.size()-1; i >= 0; --i){
ans[i] *= R;
R *= nums[i];
}
return ans;
}
};
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