LeetCode-Product of Array Except Self

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Given an array of n integers where n > 1, nums, return an array output such that output[i] is equal to the product of all the elements of nums except nums[i].

Solve it without division and in O(n).

For example, given [1,2,3,4], return [24,12,8,6].

Follow up:

Could you solve it with constant space complexity? (Note: The output array does not count as extra space for the purpose of space complexity analysis.)

自己没倒腾出来，确实挺难想的，但是当得知其中的奥秘时，不时感觉此算法真是妙。

题目大致意思就是求一个output数组，output[i]为数组nums数组除nums[i]之外数字的所有乘积，比如：

Input : [1, 2, 3, 4, 5]
Output: [(2*3*4*5), (1*3*4*5), (1*2*4*5), (1*2*3*5), (1*2*3*4)]
      = [120, 60, 40, 30, 24]

解题思路基于以下集合

{              1,         a[0],    a[0]*a[1],    a[0]*a[1]*a[2],  }
{ a[1]*a[2]*a[3],    a[2]*a[3],         a[3],                 1,  }

空间复杂度为O(n)的实现：

public int[] productExceptSelf(int[] nums) {
        int[] productBelow = new int[nums.length];
        int n = 1;
        for (int i = 0; i < nums.length; i++) {
        	productBelow[i] = n;
        	n *= nums[i];
        }
        
        int[] productAbove = new int[nums.length];
        n = 1;
        for (int i = nums.length-1; i >= 0; i--) {
        	productAbove[i] = n;
        	n *= nums[i];
        }
        
        int[] output = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
        	output[i] = productAbove[i] * productBelow[i];
        }
        return output;
    }

int[] output = new int[nums.length];
        int n = 1;
        for (int i = 0; i < nums.length; i++) {
        	output[i] = n;
        	n *= nums[i];
        }
        
        n = 1;
        for (int i = nums.length-1; i >= 0; i--) {
        	output[i] *= n;
        	n *= nums[i];
        }
        return output;

LeetCode-Product of Array Except Self

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原文地址：http://blog.csdn.net/my_jobs/article/details/47165619