#leetcode#Product of Array Except Self

最新推荐文章于 2023-02-19 11:29:03 发布

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本文介绍了一种解决给定整数数组中每个元素的乘积，但不使用除法的方法。通过创建两个数组分别存储当前位置左侧和右侧的所有元素乘积，最终通过这两个数组相乘得到结果数组。此方法适用于任何大于1的整数数组。

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Product of Array Except Self

Total Accepted: 442 Total Submissions: 1138 My Submissions

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.)

sdf

前几天刚看facebook的面经出现这题， leetocde就加上了，不能用除法，则维护当前元素左边所有元素的乘积以及右边所有元素的乘积，相乘得到 product of array except self !

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}, thenleftProd will be {1, 2, 8, 24} , and rightProd will be {72, 18, 6, 1}.

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