238. Product of Array Except Self

🟠   🔖  数组 前缀和  🔗 LeetCode

题目

Given an integer array nums, return an array answer such thatanswer[i] is equal to the product of all the elements of nums exceptnums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

Example 1:

Input: nums = [1,2,3,4]

Output: [24,12,8,6]

Example 2:

Input: nums = [-1,1,0,-3,3]

Output: [0,0,9,0,0]

Constraints:

• 2 <= nums.length <= 10^5
• -30 <= nums[i] <= 30
• The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)

题目大意

给定一个数组 nums。要求返回数组 answer，其中 answer[i] 等于 nums 中除 nums[i] 之外其余各元素的乘积。

解题思路

两次遍历

构造一个答案数组 res，长度和数组 nums 长度一致。

先从左到右遍历一遍 nums 数组，将 nums[i] 左侧的元素乘积累积起来，存储到 res 数组中。

再从右到左遍历一遍，将 nums[i] 右侧的元素乘积累积起来，再乘以原本 res[i] 的值，即为 nums 中除了 nums[i] 之外的其他所有元素乘积。

代码

// 时间复杂度 O(n) ,空间复杂度 O(1)
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var productExceptSelf = function (nums) {
  const n = nums.length;
  let res = [];

  let left = 1;
  for (let i = 0; i < n; i++) {
    res[i] = left;
    left *= nums[i];
  }

  let right = 1;
  for (let i = n - 1; i >= 0; i--) {
    res[i] *= right;
    right *= nums[i];
  }

  return res;
};

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- [42. 接雨水](https://leetcode.com/problems/trapping-rain-water)
- [152. 乘积最大子数组](./0152.md)
- [🔒 Paint House II](https://leetcode.com/problems/paint-house-ii)
- [2163. 删除元素后和的最小差值](https://leetcode.com/problems/minimum-difference-in-sums-after-removal-of-elements)
- [2906. Construct Product Matrix](https://leetcode.com/problems/construct-product-matrix)