内容简介: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:Note: Please solve it without division and in O(n).
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.)
难度: medium
题目:给定一整数组其长度大于1, 返回输出数组其元素由除当前输入元素外所有元素的乘积组成。
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 Array Except Self.
class Solution { public int[] productExceptSelf(int[] nums) { if (null == nums || nums.length < 2) { return new int[] {1}; } int n = nums.length; int[] result = new int[n]; int product = 1; for (int i = n - 1; i >= 0; i--) { result[i] = product * nums[i]; product = result[i]; } result[0] = result[1]; product = nums[0]; for (int i = 1; i < n - 1; i++) { result[i] = product * result[i + 1]; product *= nums[i]; } result[n - 1] = product; return result; } }
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