Discretization of integers

1. Introduction to Discretization:

In some algorithm problems, the numerical range will be very large, such as 0 ~ $10^9$ , but the number is relatively small, for example, only$10^5$ , but in our program, these values ​​need to be used as subscripts, and it depends on the order relationship between these values. It doesn't really matter how much it is, open a$10^{}$The array of${}^9$ is obviously unwise, because at most there is only$10^5$ data, then no matter how big the value is, we can uniquely map these numbers to 0 ~$10^{}$Among${}^5$ , we only need to open$10^{}$The array of${}^{5\; is}$ enough, and the process of mapping is discretization.

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2. Steps:

 (1)去重
 (2)二分查找映射的数

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3. Code template:
(1) De-duplication

    vector<int>alls;//假设vector中存了需要离散化的所有数据
    sort(alls.begin(),alls.end());
    alls.erase(unique(alls.begin(),alls.end()),alls.end());

( unique function )

(2) Binary search

int find(int x)
{
    
    


    int l=0;r=alls.size()-1;
    while(l<r)
    {
    
    
        int mid=l+r>>1;
        if(alls[mid]>=x)r=mid;
        else l=mid+1;
    }
    return r;//return r+1表示映射以1开始
}

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                                                                                                                                       --A certain scientific super-electromagnetic gun