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  • 238. 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.)

    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;

        }

    }

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  • 原文地址:https://www.cnblogs.com/codeskiller/p/6357310.html
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