[Leetcode] Product of Array Except Self

最新推荐文章于 2024-10-03 07:19:37 发布

Javasus

最新推荐文章于 2024-10-03 07:19:37 发布

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分类专栏： Leetcode 文章标签： leetcode

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探讨了如何在不使用除法的情况下计算数组中每个元素的乘积，除了该位置的元素本身。提供了高效的O(n)算法解决方案，并考虑了零元素的情况。

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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.)

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public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int product = 1;
        int zeroCount = 0;
        for (Integer item : nums) {
            if (item != 0) {
                product *= item;
            }
            
            if (item == 0) {
                ++zeroCount;
            }
        }
        
        
        int[] outputs = new int[nums.length];
        if (zeroCount > 1) {
            return outputs;
        }
        int zeroIndex = 0;
        for (int i = 0; i < nums.length; ++i) {
            if (zeroCount == 0) {
                outputs[i] = product / nums[i];
            } else {
                if (nums[i] != 0) {
                    outputs[i] = 0;
                } else {
                    outputs[i] = product;
                }
            }
        }
        
        return outputs;
    }
}