[leetcode]238. Product of Array Except Self

最新推荐文章于 2021-10-11 11:07:24 发布

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博客围绕给定整数数组，要求返回一个数组，其中元素等于原数组除自身外所有元素的乘积。给出了初步求解思路，即求出所有乘积和后对应位置相除，但指出输入数组含0时会出错，需进行优化。

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.)

Solution:

求出数组的所有乘积和，对应位置除一下。

class Solution {
    public int[] productExceptSelf(int[] nums) {
        if(nums == null || nums.length == 0) {
            return null;
        }
        int n = nums.length;
        int[] output = new int[n];
        int total = 1;
        for(int i = 0; i < n; i++) {
            total *= nums[i];
        }
        
        for(int j = 0; j < n; j++) {
            output[j] = total/nums[j];
        }
        return output;
    }   
}

但是有个问题，如果输入数组里面有数字为0的话，就会出错，上述程序没有考虑这种情况，优化：

class Solution {
    public int[] productExceptSelf(int[] nums) {
        if(nums == null || nums.length == 0) {
            return null;
        }
        int n = nums.length;
        int[] output = new int[n];
        int total = 1;
        for(int i = 0; i < n; i++) {
            total *= nums[i];
        }
        if(total == 0) {
            for(int k = 0; k < n; k++) {
                output[k] = calculate(nums,k);
            }
        }else{
            for(int j = 0; j < n; j++) {
                output[j] = total/nums[j];
            }
        }
        return output;
    }   
    public int calculate(int[] nums,int k) {
        int res = 1;
        for(int i = 0; i < nums.length; i++) {
            if(i == k) {
                continue;
            }else {
                res *= nums[i];
            }
        }
        return res;
    }
}