文章目录
一、题目
238. Product of Array Except Self
Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
You must write an algorithm that runs in O(n) time and without using the division operation.
Example 1:
Input: nums = [1,2,3,4]
Output: [24,12,8,6]
Example 2:
Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]
Constraints:
2 <= nums.length <= 105
-30 <= nums[i] <= 30
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)
题目来源：leetcode
二、题解
由于题目中规定不能使用除法，因此使用left和right两个数组存储nums中索引为i的元素左侧和右侧所有元素的乘积。注意vector中的初始化方法。
cpp
class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int n = nums.size();
vector<int> left(n,0);
vector<int> right(n,0);
vector<int> res;
for(int i = 0;i < n;i++){
if(i == 0) left[i] = 1;
else left[i] = left[i-1] * nums[i-1];
}
for(int i = n - 1;i >= 0;i--){
if(i == n-1) right[i] = 1;
else right[i] = right[i+1] * nums[i+1];
}
for(int i = 0;i < n;i++){
res.push_back(left[i] * right[i]);
}
return res;
}
};