LeetCode238——Product of Array Except Self

最新推荐文章于 2025-01-04 10:10:24 发布

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最新推荐文章于 2025-01-04 10:10:24 发布

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分类专栏：面试题算法题文章标签：面试题 Leetcode Array

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本文介绍了一种算法解决方案，用于在O(n)的时间复杂度内计算给定整数数组中每个元素对应的除自身外所有元素的乘积。此算法避免了使用除法操作，并且能在常数空间复杂度下完成任务。

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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.)

实现：

class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
vector<int> result;
int curr = 1;
for (int i = 0; i < nums.size(); ++i) {
result.push_back(curr);
curr = curr * nums[i];
}

curr = 1;
for (int i = nums.size() - 1; i >= 0; --i)
{
result[i] = result[i] * curr;
curr = curr * nums[i];
}
return result;
}
};