leetcode 238. Product of Array Except Self

最新推荐文章于 2021-10-11 11:07:24 发布

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最新推荐文章于 2021-10-11 11:07:24 发布

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本文介绍了一个算法问题，即计算一个整数数组中每个元素的乘积，但不包括该元素自身。文章提供了一种解决方案，考虑了数组中是否存在0的情况，并通过示例展示了算法的应用。

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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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思路：注意分情况讨论，数组里面没有0，一个0，两个以上0都是不一样的情况。

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        //test if overflow
        if(nums.length == 2){
            int temp = nums[0];
            nums[0] = nums[1];
            nums[1] = temp;
            return nums;
        }
        int tProduct = 1;
        int zcount = 0;
        boolean firstZ = true;
        for(int num : nums){
            if(num == 0 && firstZ == true){
                firstZ = false;
                zcount++;
                continue;
            }
            if(num == 0){
                zcount++;
            }
            tProduct *= num;
        }
        for(int i = 0; i < nums.length; ++i){
            if(zcount > 1){
                nums[i] = 0;
            }else if(zcount == 1){
                if(nums[i] != 0){
                    nums[i] = 0;
                }else{
                    nums[i] = tProduct;
                }
            }else if(zcount == 0){
                nums[i] = tProduct/nums[i];
            }


        }
        return nums;
    }
}