leetcode：238. Product of Array Except Self（Java）解答

Posted 2020-06-18 boy_nihao

转载请注明出处：z_zhaojun的博客
原文地址
题目地址
Product of Array Except Self

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.)

Subscribe to see which companies asked this question

解法（java）：

public int[] productExceptSelf(int[] nums) {
        // if (nums == null || nums.length < 2) {
        //     return null;
        // }
        int first = nums[0];
        int product = 1;
        int numZero = 0;
        for (int val : nums) {
            if (val == 0) {
                numZero++;
                if (numZero == 2) {
                    break;
                }
            } else {
                product *= val;
            }
        }
        nums[0] = (numZero == 2) ? 0 : (numZero == 0 ? product / first : (first == 0 ? product : 0));
        for (int i = 1; i < nums.length; i++) {
            nums[i] = (numZero == 2) ? 0 : (numZero == 0 ? product / nums[i] : (nums[i] == 0 ? product : 0));
        }
        return nums;
    }