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Product of Array Except Self

Product of Array Except Self

作者: BigBig_Fish | 来源:发表于2017-10-26 21:30 被阅读0次

    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.)

    思路

    从前向后累乘一遍,从后向前累乘一遍,一开始是用的两个数组,高分答案中遍历一遍即可。

    代码

    class Solution {
    public:
        vector<int> productExceptSelf(vector<int>& nums) {
            int len=nums.size();
            vector<int> res(len, 1);
            for(int i=1; i<len; i++){
                res[i] = nums[i-1]*res[i-1];
                res[len-i-1] = nums[len-i]*res[len-i]; 
            }
            return res;
        }
    };
    

    一遍过

    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;
        }
    };
    

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