别人的代码 Leetcode中不能用除法,我没想到做法,参考了discussion中,觉得非常巧妙 classSolution:# @param {integer[]} nums# @return {integer[]}defproductExceptSelf(self, nums): p =1n =len(nums) output = []foriinrange(0,n): output.append(p) p = p * nums[i] p =1foriinrange(n-1,-...
Can you solve this real interview question? Product of Array Except Self - Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of
LeetCode上的原题,请参见我之前的博客Product of Array Except Self。 解法一: classSolution {public:/** * @param A: Given an integers array A * @return: A long long array B and B[i]= A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1]*/vector<longlong> productExcludeItsel...
The multiplication of nums except itself is equal to the multiplication of its forward and backward. It is very similar to do the summation twice(forward and backward, but use multiplication in here) And the follow up is use a variable to store the multiplication for forward and backward. A ...
238. Product of Array Except Self 0 / 0 / 创建于 5年前 / 思路 /* The given conditions restrict serveral approach to the problem. First one disables us to calculate a product of all the numbers and divide it by each number itself. The second one disables us to have a brute force ...
We can not do one traverse to get the total product of this array and traverse to get the product except for the element itself. If one element is 0 in this array, we can not get it return value. We should analyze this problem the product except for the element itself is comprised of...
Can you solve this real interview question? Product of Array Except Self - Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of