Sword refers to the "Left Rotation String" of the Offer series

There is a shift instruction in assembly language called 循环左移(ROL). Now there is a simple task, which is to simulate the operation result of this instruction with a string. For a given character sequence S, please rotate it to the left by K bits and output the sequence. For example, the character sequence S="abcXYZdef", it is required to output the result of the circular shift left by 3 bits, that is, "XYZdefabc". Is not it simple? OK, get it done!

Method 1 : UseString

public class Solution {
    
    
    public String LeftRotateString(String str, int n) {
    
    
        int len = str.length();
        String res = "";
        if(len == 0 || str == null) return res;
        int n_true = n % len;
        for(int i = n_true; i < len; i++){
    
    
            res += str.charAt(i);
        }
        for(int i = 0; i < n_true; i++){
    
    
            res += str.charAt(i);
        }
        return res;
    }
}

In fact, it can be like this:

public class Solution {
    
    
    public String LeftRotateString(String str, int n) {
    
    
        String res = "";
        for(int i = n; i < n + str.length(); i++){
    
    
          	res += str.charAt(i % str.length());
        }
        return res;
    }
}

Time complexity: O(N)
Space complexity: O(N)

Method two : useStringBuilder

StringSplicing will produce many "fragments", useStringBuilder

public class Solution {
    
    
    public String LeftRotateString(String str, int n) {
    
    
        int len = str.length();
        String res = "";
        if(len == 0 || str == null) return res;
        int n_true = n % len;
        StringBuilder stringBuilder = new StringBuilder();
        for(int i = n; i < n + str.length(); i++){
    
    
          	stringBuilder.append(str.charAt(i % str.length()));
        }
        res = stringBuilder.toString();
        return res;
    }
}

Time complexity: O(N)
Space complexity: O(N)

Method 3 : String slicing

public class Solution {
    
    
    public String LeftRotateString(String str,int n) {
    
    
        if(str.length() == 0 || str == null) return str;
        return str.substring(n, str.length()) + str.substring(0, n);
    }
}

Time complexity: O(N)
Space complexity: O(N)

Method 4 : Use the flip function

First 0 - (n-1)flip, then n-(length-1)flip, and finally 0-(length-1)flip

For example: string "abc123",n = 3

1. First 0 - 2turn the flip into "cba123";

2. Then 3-5flip turns "cba321";

3. Finally, the 0-5flip becomes "123abc";

import java.util.Arrays;

public class Solution {
    
    
    public static String LeftRotateString(String str, int n) {
    
    
        if(str.length() == 0 || str == null) return str;
        str = reverse(str, 0 , n-1);
        str = reverse(str, n , str.length()-1);
        str =reverse(str, 0 , str.length()-1);
        return str;
    }

    private static String reverse(String str, int startIndex, int endIndex){
    
    
        char[] chars = str.toCharArray();
        while(startIndex < endIndex){
    
    
            char temp = chars[startIndex];
            chars[startIndex] = chars[endIndex];
            chars[endIndex] = temp;
            startIndex++;
            endIndex--;
        }
        str = new String(chars);
        return str;
    }
}

Time complexity: O(N)
Space complexity: O(N)