def productExceptSelf(nums): n = len(nums) res = [1]*n ## left product lp = 1 for i in range(n): res[i]=lp lp = lp*nums[i] ## right product rp = 1 for i in range(n-1, -1, -1): res[i] = res[i] * rp rp = rp*nums[i] return res 举例说明 为了更清楚地解释...
Description 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]…
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.) 题目标签:Array 题目给了我们一个 nums array, 让我们返回一个新的output array, 每一个项都是 整个数组除了自己项的乘积。题目规定了不能用除法,...
voidmultiply(constvector<double>& array1, vector<double>&array2) {intlength1 =array1.size();intlength2 =array2.size();if(length1 == length2 && length2 >1) { array2[0] =1;//计算并存储第一部分for(inti =1; i < length1; ++i) { array2[i]= array2[i -1] * array1[i -1]...
* [238] Product of Array Except Self */// @lc code=startclassSolution{public:// 不能用除法,O(N)时间,O(1)空间(返回数组除外)vector<int>productExceptSelf(vector<int>&nums){intN=nums.size();vector<int>ans(N,1);ints=1,t=1;for(inti=0;i<N;i++){ans[i]*=s;ans[N-1-i]*=t...
代码: class Solution(object): def productExceptSelf(self, nums): n = len(nums) res = [1] * n temp = 1 for i in range(0, n - 1): res[i + 1] = res[i] * nums[i] for i in range(n - 1, 0, -1): res[i] *= temp ...
LeetCode Top 100 Liked Questions 238. Product of Array Except Self (Java版; Medium) 题目描述 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]. ...
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
Given an array nums ofnintegers wheren> 1, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i]. ...
思路:先用输出数组从右向左计算累积元素的乘积,然后从左向左计算nums[i] = product(0, i-1) * result(i + 1) Runtime: 1 ms, faster than 100.00% of Java online submissions for Product of Array Except Self. Memory Usage: 40.8 MB, less than 84.97% of Java online submissions for Product of...