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));
int *re = (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++)\
{
re[i] = tmp*arr[i];
tmp*=nums[i];
}
*returnSize = numsSize;
return re;
}
238. Product of Array Except Self
原文：http://www.cnblogs.com/Alex0111/p/5386009.html