[LeetCode] Product of Array Except Self-优快云博客

本文介绍了一种高效算法，该算法能在O(n)的时间复杂度内计算出给定数组中每个元素对应的除自身外的乘积，并且避免了使用除法操作。通过两次遍历数组，先计算左侧所有元素的乘积，然后计算右侧所有元素的乘积，最终得到每个位置上的结果。

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.)

解题思路

两次遍历，第一次遍历，将0到i-1的乘积放入res[i]；第二次遍历，记录i+1到len-1的乘积，再与左边的乘积相乘，得到最终结果放入res[i]。

实现代码

// Runtime: 3 ms
public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] res = new int[nums.length];
        Arrays.fill(res, 1);
        int left = 1;
        for (int i = 0; i < nums.length - 1; i++) {
            left *= nums[i];
            res[i + 1] = left;
        }

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

        return res;
    }
}