238. Product of Array Except Self

最新推荐文章于 2024-01-03 13:51:22 发布

无名_1989

最新推荐文章于 2024-01-03 13:51:22 发布

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本文链接：https://blog.youkuaiyun.com/dongyu_1989/article/details/80474971

本文介绍了一种高效算法，用于计算一个整数数组中每个元素对应的除了自身以外所有元素的乘积，且不使用除法操作。该算法通过分别从前向后和从后向前计算累积乘积来实现，并在 O(n) 时间复杂度内完成，同时保持常数级别的空间复杂度。

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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.)

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int n = nums.size();
        int left = 1;
        int right = 1;
        vector<int> result(n, 1);
        for(int i=0; i<n; i++)
        {
            result[i]*=left;
            result[n-1-i]*=right;
            left*=nums[i];
            right*=nums[n-1-i];
        }
        return result;
    
    }
};