跳至主要內容

238. 除自身以外数组的乘积


238. 除自身以外数组的乘积

🟠   🔖  数组 前缀和  🔗 力扣open in new window LeetCodeopen in new window

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。

复杂度分析

  • 时间复杂度: O(n),其中 n 是数组的长度,需要遍历数组两次。
  • 空间复杂度: O(n),使用了一个数组来存放最终的结果。

代码

/**
 * @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;
};

相关题目

题号标题题解标签难度力扣
42接雨水[✓] 数组 双指针 2+🔴🀄️open in new window 🔗open in new window
152乘积最大子数组[✓]数组 动态规划🟠🀄️open in new window 🔗open in new window
265粉刷房子 II 🔒数组 动态规划🔴🀄️open in new window 🔗open in new window
2163删除元素后和的最小差值数组 动态规划 堆(优先队列)🔴🀄️open in new window 🔗open in new window
2906构造乘积矩阵数组 矩阵 前缀和🟠🀄️open in new window 🔗open in new window