Algorithm: minimum representation of a string

The minimum representation of a string is that the last character of a string can be put in the first place, and so on. There are N types of deformation, where N is the length of the string.

For example:

S = "00ab"

Deformation with (Omitted quotation marks) b00a ab00 0ab0

Four Types

Find the smallest Lexicographic Order, and use this algorithm.

Define three pointers, I, J, K

Initial I = 0; j = 1; k = 0

First, if s [I] <s [J], it is obvious that J ++

If s [I]> S [J], it is obvious that I = J ++

If s [I] = s [J.

In this case, all the characters between I and j must be greater than or equal to s [I ].

The other K is 0, and the first s [I + k] is searched cyclically. = S [J + k] Location

If s [I + k] <s [J + k] Then J + = k + 1

Why?

First s [I] to s [I + k-1] must be greater than or equal to s [I], because if one of the numbers is less than s [I], so this number must exist in S [J] To s [J + k-1], and because there must be one in the back, so if s [J] first met, then it will not continue to the K position, so there must be no character smaller than s [I.

Therefore, starting from any character in the sequence as the starting point will not be smaller than the current one. Therefore, it may be the minimum value only when the character after the selected sequence starts.

So J + = k + 1

If a number in the sequence is equal to the number of S [I], it must start before or after this position, so you do not need to start from this position.

Here I and j are equivalent. I have the same results before J, So I and j have the same processing. I will not explain it carefully, directly paste the Code:

In addition, if I = J, then let J ++ return to the original state.

In the end, it must be a small one that won't be moved, but the big one will keep moving backward. Therefore, you only need to output the smallest one of I and J.

int getmin(char *s){    int n=strlen(s);    int i=0,j=1,k=0,t;    while(i<n && j<n && k<n){        t=s[(i+k)%n]-s[(j+k)%n];        if (!t) k++;        else{            if (t>0) i+=k+1;             else j+=k+1;            if (i==j) j++;            k=0;        }    }    return i<j?i:j;}

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Algorithm: minimum representation of a string