linear-list/array/product-of-array-except-self

Product of Array Except Self

描述

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.)

分析

我们以一个 4 个元素的数组为例，nums=[a1,a2,a3,a4]，要想在O(n)的时间内输出结果，比较好的解决方法是提前构造好两个数组：

• [1, a1, a1*a2, a1*a2*a3]
• [a2*a3*a4, a3*a4, a4, 1]

然后两个数组一一对应相乘，即可得到最终结果 [a2*a3*a4, a1*a3*a4, a1*a2*a4, a1*a2*a3]。

不过，上述方法的空间复杂度为O(n)，可以进一步优化成常数空间，即用一个整数代替第二个数组。

代码 1 O(n)空间

// Product of Array Except Self
// Time Complexity: O(n), Space Complexity: O(n)
public class Solution {
    public int[] productExceptSelf(int[] nums) {
        final int[] result = new int[nums.length];
        final int[] left = new int[nums.length];
        final int[] right = new int[nums.length];

        left[0] = 1;
        right[nums.length - 1] = 1;

        for (int i = 1; i < nums.length; ++i) {
            left[i] = nums[i - 1] * left[i - 1];
        }

        for (int i = nums.length - 2; i >= 0; --i) {
            right[i] = nums[i + 1] * right[i + 1];
        }

        for (int i = 0; i < nums.length; ++i) {
            result[i] = left[i] * right[i];
        }

        return result;
    }
}

代码 2 O(1)空间

// Product of Array Except Self
// Time Complexity: O(n), Space Complexity: O(1)
public class Solution {
    public int[] productExceptSelf(int[] nums) {
        final int[] left = new int[nums.length];
        left[0] = 1;

        for (int i = 1; i < nums.length; ++i) {
            left[i] = nums[i - 1] * left[i - 1];
        }

        int right = 1;
        for (int i = nums.length - 1; i >= 0; --i) {
            left[i] *= right;
            right *= nums[i];
        }
        return left;
    }
}