【leetcode81】Product of Array Except Self

標(biāo)簽：

Java Linux 數(shù)據(jù)結(jié)構(gòu)

题目描述：

给定一个长度为n的整数数组Array【】，输出一个等长的数组result【】，这个输出数组，对应位置i是除了Array【i】之外，其他的所有元素的乘积

例如：

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

要求：

时间复杂度是o（n）

原文描述：

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.)

思路1：

考虑构造两个数组相乘来解决
例如nums=[a1,a2,a3,a4]，构造的数组是：
[1, a1, a1a2, a1a2a3]
[a2a3a4, a3a4, a4, 1]
上面的数组相乘，得到[a2a3a4, a1a3a4, a1a2a4, a1a2a3]

public class Solution {
    public int[] productExceptSelf(int[] nums) {
        final int[] result = new int[nums.length];
        final int[] left = new int[nums.length];
        final int[] right = new int[nums.length];

        left[0] = 1;
        right[nums.length - 1] = 1;

        for (int i = 1; i < nums.length; ++i) {
            left[i] = nums[i - 1] * left[i - 1];
        }

        for (int i = nums.length - 2; i >= 0; --i) {
            right[i] = nums[i + 1] * right[i + 1];
        }

        for (int i = 0; i < nums.length; ++i) {
            result[i] = left[i] * right[i];
        }

        return result;
    }
}

思路2：（空间复杂度o（1））

考虑上面的第二个数组的数据用一个常数代替，然后输出的数组，是不算空间的

代码：

public class Solution {
    public int[] productExceptSelf(int[] nums) {
         final int[] left = new int[nums.length];
        left[0] = 1;

        for (int i = 1; i < nums.length; ++i) {
            left[i] = nums[i - 1] * left[i - 1];
        }

        int right = 1;
        for (int i = nums.length - 1; i >= 0; --i) {
            left[i] *= right;
            right *= nums[i];
        }
        return left;
    }
}

参考：
https://soulmachine.gitbooks.io/algorithm-essentials/content/java/linear-list/array/product-of-array-except-self.html

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