238. Product of Array Except Self

 

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 numsexcept 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:

Trapping Rain Water

Maximum Product Subarray

Paint House II 

 

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

https://leetcode.com/problems/product-of-array-except-self/discuss/65622/Simple-Java-solution-in-O(n)-without-extra-space

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation