238. Product of Array Except Self

难度: Medium

题目描述

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 input is generated such that answer[i] 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.)

解题思路

代码实现

解决方案

java

 

class Solution {
    public int[] productExceptSelf(int[] nums) {

        int[] leftSum = new int[nums.length];
        int[] rightSum = new int[nums.length];
        int multiAnswer = 1;
        //从左到右乘
        for (int i = 0; i < nums.length; i++) {
            multiAnswer *= nums[i];
            leftSum[i] = multiAnswer;
        }
        //从右到左乘
        multiAnswer = 1;
        for (int i = nums.length - 1; i > -1; i--) {
            multiAnswer *= nums[i];
            rightSum[nums.length-i-1] = multiAnswer;
        }

        int[] res = new int[nums.length];
         for (int i = 0; i < nums.length; i++) {
            if(i==0){
                res[i] = rightSum[nums.length-2];
                continue;
            }
            if(i==nums.length-1){
                 res[i] = leftSum[i-1];
                continue;
            }
            res[i] = leftSum[i-1]*rightSum[nums.length-i-2];
           
        }
        return res;

    }

   
}