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) { int n = nums.size(); vector<int> result(n, 0); if(n == 0) return result; int all = 1; //nums中所有数字的乘积 int cnt = 0; //nums中0的个数 int index = -1; //nums中哪个数字是0 for(int i=0; i<n; i++) { if(nums[i] != 0) { all *= nums[i]; } else { cnt++; index = i; if(cnt == 2) return result; } } if(cnt == 2) //nums中有2个0 { return result; } else if(cnt == 1) //nums中有1个0 { result[index] = all; return result; } //正常情况 for(int i=0; i<n; i++) { if(i != index) result[i] = all / nums[i]; } return result; }};新闻热点
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