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leetcode - Product of Array Except Self
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.)
class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { vector<int> totalProductForward(nums.size()); totalProductForward[0] = nums[0]; for(int i = 1; i < nums.size(); i++){ totalProductForward[i]= totalProductForward[i-1]*nums[i]; } vector<int> totalProductBackward(nums.size()); totalProductBackward[nums.size()-1] = nums[nums.size()-1]; for(int j = nums.size()-2; j >= 0; j--){ totalProductBackward[j] = totalProductBackward[j+1]*nums[j]; } vector<int> res(nums.size()); res[0] = totalProductBackward[1]; res[nums.size()-1] = totalProductForward[nums.size()-2]; for(int j = 1; j < nums.size()-1; j++){ res[j] = (totalProductBackward[j+1] * totalProductForward[j-1]); } return res; } };
空间复杂度O(3n).分别向后和向前遍历一遍，求出累乘的积的数组，然后每个结果都可以由前一部分积和后一部分积相乘获得。
leetcode - Product of Array Except Self
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原文地址：http://www.cnblogs.com/shnj/p/4853557.html
