[LeetCode] 238. Product of Array Except Self 除本身之外的数组之积
Given an array nums
of n integers where n > 1, return an array output
such that output[i]
is equal to the product of all the elements of nums
except nums[i]
.
Example:
Input:[1,2,3,4]
Output:[24,12,8,6]
Note: Please solve it without division and in O(n).
Follow up:
Could you solve it with constant space complexity? (The output array does not count as extra space for the purpose of space complexity analysis.)
这道题给定我们一个数组,让我们返回一个新数组,对于每一个位置上的数是其他位置上数的乘积,并且限定了时间复杂度 O(n),并且不让我们用除法。如果让用除法的话,那这道题就应该属于 Easy,因为可以先遍历一遍数组求出所有数字之积,然后除以对应位置的上的数字。但是这道题禁止我们使用除法,那么我们只能另辟蹊径。我们想,对于某一个数字,如果我们知道其前面所有数字的乘积,同时也知道后面所有的数乘积,那么二者相乘就是我们要的结果,所以我们只要分别创建出这两个数组即可,分别从数组的两个方向遍历就可以分别创建出乘积累积数组。参见代码如下:
C++ 解法一:
class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { int n = nums.size(); vector<int> fwd(n, 1), bwd(n, 1), res(n); for (int i = 0; i < n - 1; ++i) { fwd[i + 1] = fwd[i] * nums[i]; } for (int i = n - 1; i > 0; --i) { bwd[i - 1] = bwd[i] * nums[i]; } for (int i = 0; i < n; ++i) { res[i] = fwd[i] * bwd[i]; } return res; } };
Java 解法一:
public class Solution { public int[] productExceptSelf(int[] nums) { int n = nums.length; int[] res = new int[n]; int[] fwd = new int[n], bwd = new int[n]; fwd[0] = 1; bwd[n - 1] = 1; for (int i = 1; i < n; ++i) { fwd[i] = fwd[i - 1] * nums[i - 1]; } for (int i = n - 2; i >= 0; --i) { bwd[i] = bwd[i + 1] * nums[i + 1]; } for (int i = 0; i < n; ++i) { res[i] = fwd[i] * bwd[i]; } return res; } }
我们可以对上面的方法进行空间上的优化,由于最终的结果都是要乘到结果 res 中,所以可以不用单独的数组来保存乘积,而是直接累积到结果 res 中,我们先从前面遍历一遍,将乘积的累积存入结果 res 中,然后从后面开始遍历,用到一个临时变量 right,初始化为1,然后每次不断累积,最终得到正确结果,参见代码如下:
C++ 解法二:
class Solution { public: vector<int> productExceptSelf(vector<int>& nums) { vector<int> res(nums.size(), 1); for (int i = 1; i < nums.size(); ++i) { res[i] = res[i - 1] * nums[i - 1]; } int right = 1; for (int i = nums.size() - 1; i >= 0; --i) { res[i] *= right; right *= nums[i]; } return res; } };
Java 解法二:
public class Solution { public int[] productExceptSelf(int[] nums) { int n = nums.length, right = 1; int[] res = new int[n]; res[0] = 1; for (int i = 1; i < n; ++i) { res[i] = res[i - 1] * nums[i - 1]; } for (int i = n - 1; i >= 0; --i) { res[i] *= right; right *= nums[i]; } return res; } }
Github 同步地址:
https://github.com/grandyang/leetcode/issues/238
类似题目:
参考资料:
https://leetcode.com/problems/product-of-array-except-self/
https://leetcode.com/problems/product-of-array-except-self/discuss/65638/My-simple-Java-solution
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