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.)
C Solution:
/**
* 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 *res = malloc(numsSize * sizeof(int));
*returnSize = numsSize;
res[0] = nums[0];
int i;
for (i = 1; i < numsSize; i++) res[i] = res[i - 1] * nums[i];
int prod = 1;
for (i = numsSize - 1; i > 0; i--) {
res[i] = res[i - 1] * prod;
prod *= nums[i];
}
res[0] = prod;
return res;
}
Python Solution:
class Solution(object):
def productExceptSelf(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
res = [nums[0]]
for i in range(1, len(nums)):
res.append(nums[i] * res[-1])
post = 1
for i in range(1, len(nums)):
res[-i] = res[-i - 1] * post
post *= nums[-i]
res[0] = post
return res
Summary:
- two arrays, two direction. funny.
LeetCode: 238. Product of Array Except Self
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