String Loop left Shift

Title Description: Given a string s[0...n-1], required to move the first k characters of s to the tail of s, such as the string "ABCdef" before the 2 characters ' a ', ' B ' moved to the end of the string, get the new string "Cdefab": That is, the string loop left K.

The algorithm requires: The time complexity is O (n), the space complexity is O (1).

Problem Analysis:

1. The Act of Violent displacement

Each loop moves left 1 bits, calls K times, Time complexity O (kn), Space complexity O (1)

2. Open an array of k+1 lengths and make three copies

S[0...K]→T[0...K]

S[K+1...N-1]→S[0...N-K-1]
T[0...k]→s[n-k ... N-1]
Time complexity O (n), spatial complexity O (k)

None of the two are compliant with algorithmic requirements.

In fact, there is a kind of algorithm that satisfies the condition:

(X ' Y ') ' =yx

such as: abcdef

X=ab X ' =ba
Y=cdef Y ' =fedc
(X ' Y ') ' = (BAFEDC) ' =cdefab

Time complexity O (N), Spatial complexity O (1)

Code implementation:

  1#include <iostream>2#include <string>3 using namespacestd; 4   5 voidReverseString (string&AMP;STR,intMintN)6 {  7      for(inti = m; I < n; ++i)8     {  9         Chartemp =Str[i];TenStr[i] =Str[n]; OneStr[n] =temp; A--N; -     }  - }  the   - intMainintargcConst Char*argv[]) - {  -     stringstr ="abcdef";  +     intLen =str.size (); -     intK =2;//Loop left shift 2 bit  +K%= Len;//prevent cross-border  AReverseString (str,0K1);  atReverseString (str, k, Len-1);  -ReverseString (str,0, Len-1);  -cout << str <<Endl; -     return 0;  -}

Results:

PS:STL provides a reverse flip string function

String loop shift left