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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最新推荐文章于 2021-10-11 11:07:24 发布

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分类专栏： Leetcode Array

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本文介绍了一种高效算法，用于求解数组中每个元素等于该数组其他所有元素的乘积的问题，避免了使用除法操作。通过两次遍历数组分别计算左侧乘积和右侧乘积来实现。

http://stackoverflow.com/questions/2680548/given-an-array-of-numbers-return-array-of-products-of-all-other-numbers-no-div

/**
 * The product of an index equals to the product left to the indx times the product right to the index.
 * e.g. [1, 2, 3, 4]
 * Calculate the product left to an index i (from left to right)
 * A = [1, 1, 2, 6] the product left to index 0 is 1
 * Calculate the product right to an index i (from right to left)
 * B = [24, 12, 4, 1] 
 * The result would be 
 * A * B = [24, 12, 8, 6]
 */ 
public class Solution {
    public int[] productExceptSelf(int[] nums) {
        int[] res = new int[nums.length];
        res[0] = 1;
        
        // calculate the left product 
        for (int i=1; i<nums.length; i++)
            res[i] = nums[i-1] * res[i-1];
        System.out.println(res[1]);
        // use a variable to save the right product
        int rp = 1; 
        for (int i=nums.length-1; i>-1; i--) {
            res[i] *= rp;
            rp     *= nums[i];
        }
        
        return res;
    }
}