[LeetCode 238] Product of Array Except Self

标签：leetcode   math   

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.)

Solution:

straightforward

1. there would be several 0s in the array. if have more than one 0, then result should be [0000], if only has one 0, result [0000{non-zero}000]

2. if no 0s in array, keep all element product, each time divided by array[i]

public int[] productExceptSelf(int[] nums) {
        int[] result = new int[nums.length];
        long pro = 1;
        int zeroNum = 0;
        for(int i=0;i<nums.length;i++){
            result[i] = 0;
            if(nums[i] == 0) {
                zeroNum++;
            }else {
                pro *= (long) nums[i];
            }
        }
        if(zeroNum>1) return result;
        for(int i=0;i<nums.length;i++){
            if(nums[i] == 0) {
                result[i] = (int)pro;
            }else {
                result[i] = (zeroNum == 1) ? 0 : (int) (pro/nums[i]);
            }
        }
        return result;
    }

[LeetCode 238] Product of Array Except Self

标签：leetcode   math   

原文地址：http://blog.csdn.net/sbitswc/article/details/48514873