wk2 238. Product of Array Except Self

原创于 2016-09-14 01:17:08 发布 · 246 阅读

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本文介绍一种高效算法，解决求解数组中每个元素对应除自身外其他元素乘积的问题，采用双遍历策略，避免了使用除法运算，实现了O(n)的时间复杂度。

Difficulty: Medium

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.)

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int s = nums.size(), i = 0;
        vector<int> <code>output</code>(s, 1);
        for (i = 1; i < s; ++i)
            <code>output</code>[i] = <code>output</code>[i-1] * nums[i-1];
        int k = nums[s-1];
        for (i = s - 2; i >=0; --i)
        {
            <code>output</code>[i] *= k;
            k *= nums[i];
        }
        return <code>output</code>;
    }
};

思路是先计算output[i]左边所有数的乘积，再计算output[i]右边所有数的乘积，用两个for循环来完成计算。