[Leetcode] Product of Array Except self

Given an array nums of n integers where n > 1, return a array output such that's equal to output[i] th E product of all the elements of nums except nums[i] .

Example:

Input:  [1,2,3,4][24,12,8,6]

Note:please solve it without division and in O (n).

Follow up:
Could solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)

Analysis: The topic of translation: the need to calculate an array, in addition to the product of all elements other than itself, can not use division.      Looking at the requirements, without looking at the limit, we have two ideas: 1. Calculate the product of all the elements, then traverse from beginning to end, for any i,res[i] = product/nums[i]. This is, of course, the most general approach, but it is not possible to use division, and this method is ignored. 2. Join us to calculate the results of I position res[i], is actually calculated res[i]=nums[0]*nums[1]*...*nums[i-1]*nums[i+1]*...*nums[nums.length-1] The following is metaphysics: Res[i] =nums[0]*nums[1]*...*nums[i-1]*nums[i+1]*...*nums[nums.length-1]= (Nums[0]*nums[1]*...*nums[i-1]) * (nums[i+1]*     *NUMS[NUMS.LENGTH-1]) that is to divide the result into two parts, left and right, so res[i]=left[i]*right[i], here is how to determine the left and right arrays. Left[i]=nums[0]*nums[1]*...*nums[i-1], this is a recursion. , Left[i]=left[i-1]*nums[i-1] code is as follows:

1 classSolution {2      Public int[] Productexceptself (int[] nums) {3         intn =nums.length;4 5         int[] left =New int[n];6Left[0] = 1;7          for(inti = 1; I < n; i + + )8Left[i] = left[i-1] * nums[i-1];9 Ten         int[] right =New int[n]; OneRight[n-1] = 1; A          for(inti = n-2; I >= 0; I-- ) -Right[i] = right[i+1] * nums[i+1]; -          the         int[] res =New int[n]; -          for(inti = 0; I < n; i + + ) -Res[i] = left[i] *Right[i]; -          +         returnRes; -     } +}

Run time 1ms, beat 100%.

But this method still uses the extra space, the topic said hoped can not use the extra space. Therefore, you can use a right variable instead of the one that is left.

Refer to the code of the discuss great God:

1  Public int[] Productexceptself (int[] nums) {2     intn =nums.length;3     int[] res =New int[n];4Res[0] = 1;5      for(inti = 1; I < n; i++) {6Res[i] = res[i-1] * nums[i-1];7     }8     intright = 1;9      for(inti = n-1; I >= 0; i--) {TenRes[i] *=Right ; OneRight *=Nums[i]; A     } -     returnRes; -}

[Leetcode] Product of Array Except self