238. Product of Array Except Self（对O（n）和递归又有了新的理解）

238. Product of Array Except Self

 

 

Total Accepted: 41565 Total Submissions: 97898 Difficulty: Medium

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.)

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Code:

int* productExceptSelf(int* nums, int numsSize, int* returnSize) {
    int i,tmp = 1;
    int *arr = (int *)malloc(sizeof(int)*(numsSize));
 
    arr[numsSize-1] = 1;
    for(i = numsSize-2;i>=0;i--)
        arr[i] = arr[i+1]*nums[i+1];
    for(i = 0;i<numsSize;i++)
    {
        arr[i] = tmp*arr[i];
        tmp*=nums[i];
    }
    *returnSize = numsSize;
    return arr;
}

//一个嵌套把前边的乘积之和带过来，再在跳出嵌套之前得出相应的结果。

int* productExceptSelf(int* nums, int numsSize, int* returnSize) {
        multiply(nums, 1, 0, numsSize);
        *returnSize = numsSize;
        return nums;
    }

    int multiply(int *a, int fwdProduct, int indx, int N) {
        int revProduct = 1;
        if (indx < N) {
            revProduct = multiply(a, fwdProduct * a[indx], indx + 1, N);
            int cur = a[indx];
            a[indx] = fwdProduct * revProduct;
            revProduct *= cur;
        }
        return revProduct;
    }