238. Product of Array Except Self

Posted 2020-08-29 会咬人的兔子

238. Product of Array Except Self

• Total Accepted: 80998
• Total Submissions: 171009
• Difficulty: Medium
• Contributors: Admin

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.)

Solution1:

Use two vectors to record the product of numbers befor/after the current number.

The result[i] is the product of  fwd[i] * bwd[i]

 1 class Solution {
 2 public:
 3     vector<int> productExceptSelf(vector<int>& nums) {
 4         int n = nums.size();
 5         vector<int> fwd(n, 1);
 6         vector<int> bwd(n, 1);
 7         vector<int> res(n);
 8         //forward
 9         for(int i = 0; i < n - 1; i++){
10             fwd[i + 1] = fwd[i] * nums[i];
11         }
12         //backward
13         for(int i = n - 1; i > 0; i--){
14             bwd[i - 1] = bwd[i] * nums[i];
15         }
16         for(int i = 0; i < n; i++){
17             res[i] = fwd[i] * bwd[i];
18         }
19         return res;
20     }
21 };

Solution 2:

我们可以对上面的方法进行空间上的优化，由于最终的结果都是要乘到结果res中，所以我们可以不用单独的数组来保存乘积，而是直接累积到res中，我们先从前面遍历一遍，将乘积的累积存入res中，然后从后面开始遍历，用到一个临时变量right，初始化为1，然后每次不断累积，最终得到正确结果

 1 class Solution {
 2 public:
 3     vector<int> productExceptSelf(vector<int>& nums) {
 4         vector<int> res(nums.size(), 1);
 5         for (int i = 1; i < nums.size(); ++i) {
 6             res[i] = res[i - 1] * nums[i - 1];
 7         }
 8         int right = 1;
 9         for (int i = nums.size() - 1; i >= 0; --i) {
10             res[i] *= right;
11             right *= nums[i];
12         }
13         return res;
14     }
15 };