Given an array nums of n integers where n > 1, return an array output such that output[i] is equal
to the product of all the elements of nums except nums[i].
Example:
Input: [1,2,3,4]
Output: [24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity?
(The output array does not count as extra space for the purpose of space complexity analysis.)
class Solution:
def productExceptSelf(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
n = len(nums)
res = [None]*n
res[0] = 1
for i in range(1,n):
res[i] = res[i-1] * nums[i-1]
right = 1
for i in range(n-1,-1,-1):
res[i] *= right
right *= nums[i]
return res;
这个题并不难，开始我尝试了一下算所有数字的product然后再遍历整个数组，输出product/当前数字。 后来发现这样做一旦遇到0 基本属于GG 因为所有数字的成乘积就变成了0 处理起来很麻烦 需要很多判断，因为整个数组可能在多个位置出现0。
这个解法的精髓就在于，把本位置以及以前的数字成积算出来放在res[]这个list 里，第二遍遍历的时候从后往前用同样的方法，只不过再乘之前算好 的res 中的数值
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