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| 1 | +/* |
| 2 | + * @lc app=leetcode.cn id=238 lang=cpp |
| 3 | + * |
| 4 | + * [238] 除自身以外数组的乘积 |
| 5 | + * |
| 6 | + * https://leetcode.cn/problems/product-of-array-except-self/description/ |
| 7 | + * |
| 8 | + * algorithms |
| 9 | + * Medium (74.80%) |
| 10 | + * Likes: 1457 |
| 11 | + * Dislikes: 0 |
| 12 | + * Total Accepted: 270.8K |
| 13 | + * Total Submissions: 362K |
| 14 | + * Testcase Example: '[1,2,3,4]' |
| 15 | + * |
| 16 | + * 给你一个整数数组 nums,返回 |
| 17 | + * 数组 answer ,其中 answer[i] 等于 nums 中除 nums[i] 之外其余各元素的乘积 。 |
| 18 | + * |
| 19 | + * 题目数据 保证 数组 nums之中任意元素的全部前缀元素和后缀的乘积都在 32 位 |
| 20 | + * 整数范围内。 |
| 21 | + * |
| 22 | + * 请不要使用除法,且在 O(n) 时间复杂度内完成此题。 |
| 23 | + * |
| 24 | + * |
| 25 | + * |
| 26 | + * 示例 1: |
| 27 | + * |
| 28 | + * |
| 29 | + * 输入: nums = [1,2,3,4] |
| 30 | + * 输出: [24,12,8,6] |
| 31 | + * |
| 32 | + * |
| 33 | + * 示例 2: |
| 34 | + * |
| 35 | + * |
| 36 | + * 输入: nums = [-1,1,0,-3,3] |
| 37 | + * 输出: [0,0,9,0,0] |
| 38 | + * |
| 39 | + * |
| 40 | + * |
| 41 | + * |
| 42 | + * 提示: |
| 43 | + * |
| 44 | + * |
| 45 | + * 2 <= nums.length <= 10^5 |
| 46 | + * -30 <= nums[i] <= 30 |
| 47 | + * 保证 数组 nums之中任意元素的全部前缀元素和后缀的乘积都在 32 位 整数范围内 |
| 48 | + * |
| 49 | + * |
| 50 | + * |
| 51 | + * |
| 52 | + * 进阶:你可以在 O(1) 的额外空间复杂度内完成这个题目吗?( |
| 53 | + * 出于对空间复杂度分析的目的,输出数组不被视为额外空间。) |
| 54 | + * |
| 55 | + */ |
| 56 | + |
| 57 | +#include <vector> |
| 58 | +using namespace std; |
| 59 | + |
| 60 | +// @lc code=start |
| 61 | +class Solution { |
| 62 | +public: |
| 63 | + // 左右两个数组,分别代表当前元素左侧乘积和右侧乘积,两者想乘即为当前位置结果 |
| 64 | + vector<int> productExceptSelf(vector<int> &nums) { |
| 65 | + int n = nums.size(); |
| 66 | + if (nums.empty()) { |
| 67 | + return vector<int>(); |
| 68 | + } |
| 69 | + vector<int> left(n), right(n); |
| 70 | + left[0] = 1, right[n - 1] = 1; |
| 71 | + for (int i = 1; i < n; i++) { |
| 72 | + left[i] = left[i - 1] * nums[i - 1]; |
| 73 | + } |
| 74 | + for (int i = n - 2; i >= 0; i--) { |
| 75 | + right[i] = right[i + 1] * nums[i + 1]; |
| 76 | + } |
| 77 | + |
| 78 | + vector<int> ret(n); |
| 79 | + for (int i = 0; i < n; i++) { |
| 80 | + ret[i] = left[i] * right[i]; |
| 81 | + } |
| 82 | + return ret; |
| 83 | + } |
| 84 | +}; |
| 85 | +// @lc code=end |
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