（Leetcode）238. Product of Array Except Self

最新推荐文章于 2021-07-22 23:09:09 发布

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最新推荐文章于 2021-07-22 23:09:09 发布

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分类专栏： Leetcode Algorithm

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本文介绍了一种高效算法，用于求解给定数组中除自身外所有元素的乘积，该算法避免了使用除法操作，并实现了O(n)的时间复杂度。通过两个遍历过程，分别计算左侧和右侧元素的累积乘积，最终得到每个元素对应的答案。

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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.)

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        if(nums==null||nums.length==0) return nums;
        int res[] = new int[nums.length];
        int left = 1;
        for(int i=0; i<nums.length; i++){
            res[i] = left; //res[k] = nums[k-1]*nums[k-2]*...*nums[0]
            left *= nums[i];
        }
        int right = 1;
        for(int i=nums.length-1; i>=0; i--){
            res[i] *=right;
            right *=nums[i];  //res[k] = res[k] * (nums[n-1]*nums[n-2]*...*nums[k+1])
        }
        return res;
    }
}