[LeetCode] Perform String Shifts

You are given a string s containing lowercase English letters, and a matrix shift, where shift[i] = [direction, amount]:

• direction can be 0 (for left shift) or 1 (for right shift). 
• amount is the amount by which string s is to be shifted.
• A left shift by 1 means remove the first character of s and append it to the end.
• Similarly, a right shift by 1 means remove the last character of s and add it to the beginning.

Return the final string after all operations.

 

Example 1:

Input: s = "abc", shift = [[0,1],[1,2]] Output: "cab" Explanation:  [0,1] means shift to left by 1. "abc" -> "bca" [1,2] means shift to right by 2. "bca" -> "cab"

Example 2:

Input: s = "abcdefg", shift = [[1,1],[1,1],[0,2],[1,3]] Output: "efgabcd" Explanation:  [1,1] means shift to right by 1. "abcdefg" -> "gabcdef" [1,1] means shift to right by 1. "gabcdef" -> "fgabcde" [0,2] means shift to left by 2. "fgabcde" -> "abcdefg" [1,3] means shift to right by 3. "abcdefg" -> "efgabcd"

 

Constraints:

• 1 <= s.length <= 100
• s only contains lower case English letters.
• 1 <= shift.length <= 100
• shift[i].length == 2
• 0 <= shift[i][0] <= 1
• 0 <= shift[i][1] <= 100

 

#1

import java.util.ArrayDeque;

public class Solution{
	
	public String stringShift(String s, int[][] shift) {
		
		ArrayDeque<Character> deque = new ArrayDeque<>();
		String r = "";  // result
        	int n = s.length(), i, d, c;  // direction, count
        	for( i = 0; i < n; i++ )
        		deque.add( s.charAt(i) );
		for( i = 0; i < shift.length; i++ ) {
			d = shift[i][0]; c = shift[i][1];
			while( c-- > 0 ) {
				if( d == 0 )  deque.addLast( deque.pollFirst() );
				if( d == 1 )  deque.addFirst( deque.pollLast() );
			}
		}
		for( i = 0; i < n; i++ )
        		r += deque.pollFirst();
		return r;
	}
}

 

#2

public class Solution {

	public static String stringShift(String s, int[][] shift) {
    
		String o = s, r = ""; // original, result
		int d, c; // direction, count
        	int n = s.length();
        	int i, j;
		for( i = 0; i < shift.length; i++ ) {
			d = shift[i][0]; c = shift[i][1]; //r = "";
			while( c-- > 0 ) {
				r = "";
				if( d == 0 ) {
					for( j = 1; j < n; j++ ) 
						r += o.charAt(j);
					r += o.charAt(0);
				}
				if( d == 1 ) {
					r += o.charAt(n-1);
					for( j = 0; j < n-1; j++ )
						r += o.charAt(j);
				}
				o = r;	
			}
		}
		return r;
    }
    
}