[LeetCode]238. Product of Array Except Self

最新推荐文章于 2021-07-22 23:09:09 发布

转载最新推荐文章于 2021-07-22 23:09:09 发布 · 240 阅读

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本文介绍了一种高效算法，用于求解给定数组中每个元素对应的其他所有元素的乘积，且整个过程不使用除法操作。该算法通过两次遍历实现了O(n)的时间复杂度。

Medium

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].

2ms:

 public int[] productExceptSelf(int[] nums) {
    int n = nums.length;
    int[] res = new int[n];
    res[0] = 1;
    for (int i = 1; i < n; i++) {
        res[i] = res[i - 1] * nums[i - 1];
    }
    int right = 1;
    for (int i = n - 1; i >= 0; i--) {
        res[i] *= right;
        right *= nums[i];
    }
    return res;
}