Product of Array Except Self

Product of Array Except Self

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].

就是返回除开自身外，其他数的乘积

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.)

思路

恩，整体就是分成两部分，以自己为分界，前一部分，后一部分

比如A[7]
B[3]: a[0]a[1]a[2] / a[6]a[5]a[4]
B[4]: a[0]a[1]a[2]a[3] / a[6]a[5]

前一部分，正序，用到之前的结果
后一部分，逆序，用到之前的结果，但有个临时变量存储

注意事项

大概就是初值赋值1，然后for的条件

代码

/** * Return an array of size *returnSize. * Note: The returned array must be malloced, assume caller calls free(). */int* productExceptSelf(int* nums, int numsSize, int* returnSize) {    int * ret = (int *)malloc(sizeof(int)*numsSize);    if(ret)    {        ret[0] = ret[numsSize-1] = 1;        int i;        int tmp;        for(i = 1; i<numsSize;i++)        {            ret[i] = ret[i-1]*nums[i-1];        }        tmp = 1;        for(i = numsSize-1;i>=0;i--)        {            ret[i]=ret[i]*tmp;            tmp *= nums[i];        }        * returnSize = numsSize;    }    return ret;}