标签：where ret temp namespace space nts ace const 一个
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.)
注释：用两个指针，前面的指针fromBegin比返回的res少乘一个当前的i，后面的指针也是如此。当i到达结尾时，前后都覆盖到了。
#include <cstdio> #include <vector> #include<iostream> using namespace std; class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { int n=nums.size(); int fromBegin=1; int fromLast=1; vector<int> res(n,1); for(int i=0;i<n;i++){ res[i]*=fromBegin; fromBegin*=nums[i]; res[n-1-i]*=fromLast; fromLast*=nums[n-1-i]; } return res; } }; int main(){ vector<int> temp; temp.push_back(2); temp.push_back(2); temp.push_back(2); temp.push_back(2); temp.push_back(2); temp.push_back(2); temp.push_back(2); temp.push_back(2); Solution test; test.productExceptSelf(temp); return 0; }
238. [LeetCode] Product of Array Except Self
标签：where ret temp namespace space nts ace const 一个
原文地址：https://www.cnblogs.com/250101249-sxy/p/10438014.html
