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.)
此题解法是进行两次遍历，第一次从前往后遍历，第二次从后往前遍历，没遍历到一个数组值的时候，计算该数组值前遍历过的值的乘积，然后保存在乘积的数组里面，代码如下：
public class Solution {
public int[] productExceptSelf(int[] nums) {
int[] res= new int[nums.length];
int left = 1;
int right = 1;
res[0] = 1;
for(int i=1;i<nums.length;i++){
res[i] = res[i-1]*nums[i-1];
}
for(int i=nums.length-1;i>=0;i--){
res[i] = res[i]*right;
right = right*nums[i];
}
return res;
}
}
238. Product of Array Except Self
原文：http://www.cnblogs.com/codeskiller/p/6357310.html