[LeetCode] Product of Array Except Self

最新推荐文章于 2025-04-16 14:30:40 发布

xkfcl

最新推荐文章于 2025-04-16 14:30:40 发布

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分类专栏： Algorithm 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/xkfcl/article/details/49479375

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本文介绍了一种算法，该算法能在O(n)时间内求出数组中每个元素对应的除自身以外的所有元素乘积，不使用额外空间。通过双向遍历数组，分别从前向后和从后向前计算累积乘积来实现。

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This problem is easy if we can use division. The idea is get the product of all numbers and division each of them to get the result we want. Linear time and constant space.

Here we can't use this method. One intuitive thinking is: I can get the product from the first one to each number in the array and do it in the opposite direction. By observation, it's clear that by jumping one node we can get the product without it.

vector<int> productExceptSelf(vector<int>& nums) {
	int len=nums.size();
	vector<int> res;
	if(!len) return res;
vector<int> forw(1,1);
	vector<int> back(1,1);
	for(int i=0;i<len-1;i++)
	{
		forw.push_back(forw[i]*nums[i]);
		back.push_back(back[i]*nums[len-i-1]);
	}
	for(int i=0;i<len;i++)
	{
		res.push_back(forw[i]*back[len-i-1]);
	}
	return res;
	
}