238. Product of Array Except Self

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238. Product of Array Except Self

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Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of 
nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

 

Example 1:

Input: nums = [1,2,3,4]
Output: [24,12,8,6]
Example 2:

Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]
 

Constraints:

2 <= nums.length <= 105
-30 <= nums[i] <= 30
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
 

Follow up: Can you solve the problem in O(1) extra space complexity? 
(The output array does not count as extra space for space complexity analysis.)

难度 : Medium

思路

只需要用一个数组 ans,使得ans[i] = leftProd * rightProd
从左边开始扫描用一个临时变量保存从左边开始到当前位置的running prod,同时更新ans
然后从右边扫描用一个临时变量保存从右边开始到当前位置的running prod

时间复杂度: O(N)
空间复杂度: O(N)

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

        int right = 1;
        for (int i = n - 1; i >= 0; i--) {
            ans[i] *= right;
            right *= nums[i];
        }
        return ans;
    }
}