238. 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 ofnums 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.)

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数组除自身元素外,其他元素间的乘积,形成新的数组.

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思路:

设置两个整数变量,start=end=1,新的返回数组vector<int> re(nums.size(),1);

分别从数组首尾依次 对 re中当前元素相乘&赋值

遍历一边后,正好re中每个元素就是所要求的.

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代码

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        if(nums.empty()) return vector<int>();

        vector<int> re(nums.size(),1);
        int start = 1;
        int e = 1;
        for(int i =0;i<(int)nums.size();i++){
            re[i] *= start;
            start *= nums[i];
            re[nums.size()-i-1] *=e;
            e *= nums[nums.size()-i-1];
        }
        for(auto i:re) cout<<i<<" ";
        cout<<endl;
        return re;
    }
};

238. Product of Array Except Self

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原文地址：http://www.cnblogs.com/li-daphne/p/5608782.html