[LeetCode] Product of Array Except Self 除本身之外的数组之积

 

Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:
Could you solve it with constant space complexity? (Note: 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,所以我们可以不用单独的数组来保存乘积,而是直接累积到res中,我们先从前面遍历一遍,将乘积的累积存入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;
    }
}

 

参考资料:

https://discuss.leetcode.com/topic/18864/simple-java-solution-in-o-n-without-extra-space

 

    原文作者:Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/4650187.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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