【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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本文介绍了一种高效算法，用于求解给定数组中每个元素对应的除自身以外所有元素的乘积，且不使用除法操作。通过两次遍历实现O(n)时间复杂度，同时保持常数空间复杂度。

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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.)

class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] output = new int[nums.length];
        if(nums.length <= 1 || nums == null) {
            return output;
        }
        //计算下三角 up->down
        output[0] = 1;
        for(int i = 1;i < output.length;i++) {
            output[i] = output[i - 1] * nums[i - 1];
        }
        
        //计算上三角 down->up
        int temp = 1;
        for(int i = output.length - 2;i >= 0;i--) {
            temp = temp * nums[i + 1];
            output[i] = output[i] * temp;
        }
        
        return output;
    }
}