Code & Func

第43天。

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.)

这里说不能使用除法，我的想法就是自己实现一个除法:

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int Div(unsigned  a,unsigned b) {
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    int x,y;
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    int ans = 0;
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    while(a >= b) {
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        x = b;
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        y = 1;
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        while( a >= (x<<1)) {
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            x <<= 1;
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            y <<= 1;
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        }
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        a -= x;
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        ans += y;
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    }
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    return ans;
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}
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int div(int a,int b) {
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    if (a > 0 && b > 0) return Div(a,b);
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    else if (a < 0 && b < 0) return Div(-a,-b);
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    else if (a < 0) return -Div(-a,b);
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    else return -Div(a,-b);
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}

然后剩下的东西就是将0这个特例排除掉了：

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vector<int> productExceptSelf(vector<int>& nums) {
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    vector<int> ret(nums.size(),0);
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    long long product = 1;
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    int zero_count = 0;
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    for(auto i:nums)
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        if (i != 0) product*=i;
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        else zero_count++;
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    cout << zero_count << endl;
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    if (zero_count > 1) return ret;
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    if (zero_count == 1) {
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        for(int i = 0;i < nums.size();i++) {
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            if (nums[i] != 0) ret[i] = 0;
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            else ret[i] = product;
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        }
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        return ret;
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    }
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    for(int i = 0;i < ret.size();i++) {
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    //    if (nums[i] == 0) ret[i] = product[i];
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        ret[i] = div((int)product,nums[i]);
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    }
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    return ret;
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}

但是看了dicuss的做法，我感觉的理解是错的：

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def productExceptSelf(self, nums):
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        p = 1
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        n = len(nums)
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        output = []
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        for i in range(0,n):
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            output.append(p)
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            p = p * nums[i]
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        p = 1
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        for i in range(n-1,-1,-1):
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            output[i] = output[i] * p
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            p = p * nums[i]
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        return output

恩，今天写的有点急，因为我周五安全导论还要考试，然而我还一堆东西不会。。。