Leetcode Product of Array Except Self

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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.)

解题思路：

参考答案。

Because we cannot use division, so assume we have two integer arrays with the same length of nums, 

int[] leftProd = new int[nums.length]; int[] rightProd = new int[nums.length],

we store the product of all the left elements in leftProd and the product of all the right elements in rightProd,

then the product of leftProd[i] and rightProd[i] will be the value we want to put into the result.

Take the example of num[] = {2, 4, 3, 6}, then leftProd will be {1, 2, 8, 24} , and rightProd will be {72, 18, 6, 1}.

Java code:

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] result = new int[nums.length];
        for(int i = 0; i < nums.length; i++) {
            if( i == 0){
                result[i] = 1;
            }else{
                result[i] = result[i-1] * nums[i-1];
            }
        }
        int prod = 1;
        for(int i = nums.length-1; i >= 0; i--){
            result[i] *= prod;
            prod *= nums[i];
        }
        return result;
    }
}

Reference:

1. https://leetcode.com/discuss/46150/java-o-n-solution-no-extra-space-with-explanation

Leetcode Product of Array Except Self

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原文地址：http://www.cnblogs.com/anne-vista/p/4899729.html