leetcode--Product of Array Except Self
2016-03-12 21:54
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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.)
解:
这几道题都没想出来啊。。
基本思想就是先遍历一遍,用数组存储前i个元素的乘积但是往右边挫了一个,也就是不包含当前i的前i-1个元素的积,然后再从右边扫一遍,记录后几个元素乘积
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.)
解:
这几道题都没想出来啊。。
基本思想就是先遍历一遍,用数组存储前i个元素的乘积但是往右边挫了一个,也就是不包含当前i的前i-1个元素的积,然后再从右边扫一遍,记录后几个元素乘积
public class Solution { public int[] productExceptSelf(int[] nums) { int n = nums.length; int[] res = new int ; res[0] = 1; for (int i = 1; i < n; i++) { res[i] = res[i - 1] * nums[i - 1]; } int right = 1; for (int i = n - 1; i >= 0; i--) { res[i] *= right; right *= nums[i]; } return res; }
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