我们有一个方阵，它的大小不断地扩大2倍。给定矩阵中任意时刻（i，j）处的序列，我们需要返回（i，（j+N-1）%N）处的序列，其中N是矩阵的大小。 当我们说矩阵正在扩展时，扩展后的矩阵是通过将原始2×2矩阵的每个元素与当前的nxn矩阵本身相乘而形成的。扩展矩阵的维数为2N x 2N。
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For Instance, consider below 2x2 matrix, [a b] Expanding it will result in a 4x4 matrix as follows:ax[a b] bx[a b] [aa ab ba bb] [ac ad bc bd] --> [ca cb da db]cx[a b] dx[a b] [cc cd dc dd] Expanding it again results in an 8x8 matrix as follows, and so on.ax[aa ab ba bb] bx[aa ab ba bb] [aaa aab aba abb baa bab bba bbb] [ac ad bc bd] [ac ad bc bd] [aac aad abc abd bac bad bbc bbd] [ca cb da db] [ca cb da db] [aca acb ada adb bca bcb bda bdb] [cc cd dc dd] [cc cd dc dd] [acc acd adc add bcc bcd bdc bdd] --> [caa cab cba cbb daa dab dba dbb]cx[aa ab ba bb] dx[aa ab ba bb] [cac cad cbc cbd dac dad dbc dbd] [ac ad bc bd] [ac ad bc bd] [cca ccb cda cdb dca dcb dda ddb] [ca cb da db] [ca cb da db] [ccc ccd cdc cdd dcc dcd ddc ddd] [cc cd dc dd] [cc cd dc dd]
基本上，对于一个给定的序列，我们需要找出剩下的序列。矩阵可以假设为圆形，即位置（i，0）处的序列应返回位置（i，N-1）处的序列 例如：
Input: str = ddaOutput: dcbInput: str = ccaOutput: ddbInput: str = aacbddcOutput: aacbdcd
我们强烈建议您尽量减少浏览器，并先自己尝试。 如果我们仔细分析，我们可以看到一种模式。 算法： 我们从最右边的位置开始扫描字符串，并对每个字符执行以下操作- 1.如果当前字符为“b”或“d”，则分别更改为“a”或“c”，并返回字符串。 2.如果当前字符为“a”或“c”，则分别将其更改为“b”或“d”，并移至左侧的下一个字符。对下一个左字符重复步骤1。
C++
// C++ Program to return previous element in an expanding // matrix. #include <bits/stdc++.h> using namespace std; // Returns left of str in an expanding matrix of // a, b, c, and d. string findLeft(string str) { int n = str.length(); // Start from rightmost position while (n--) { // If the current character is ‘b’ or ‘d’, // change to ‘a’ or ‘c’ respectively and // break the loop if (str[n] == 'd' ) { str[n] = 'c' ; break ; } if (str[n] == 'b' ) { str[n] = 'a' ; break ; } // If the current character is ‘a’ or ‘c’, // change it to ‘b’ or ‘d’ respectively if (str[n] == 'a' ) str[n] = 'b' ; else if (str[n] == 'c' ) str[n] = 'd' ; } return str; } // driver program to test above method int main() { string str = "aacbddc" ; cout << "Left of " << str << " is " << findLeft(str); return 0; } |
JAVA
// Java program to return previous element // in an expanding matrix import java.io.*; class GFG { // Returns left of str in an expanding matrix // of a, b, c and d. static StringBuilder findLeft(StringBuilder str) { int n = str.length(); // Start from rightmost position while (n > 0 ) { n--; // If the current character is b or d, // change to a or c respectively and // break the loop if (str.charAt(n) == 'd' ) { str.setCharAt(n, 'c' ); break ; } if (str.charAt(n) == 'b' ) { str.setCharAt(n, 'a' ); break ; } // If the current character is a or c, // change it to b or d respectively if (str.charAt(n) == 'a' ) str.setCharAt(n, 'b' ); else if (str.charAt(n) == 'c' ) str.setCharAt(n, 'd' ); } return str; } // driver program to test above method public static void main (String[] args) { StringBuilder str = new StringBuilder( "aacbddc" ); System.out.print( "Left of " + str + " is " + findLeft(str)); } } // This code is contributed by Prakriti Gupta |
Python3
# Python3 Program to return previous element # in an expanding matrix. # Returns left of str in an # expanding matrix of a, b, c, and d. def findLeft( str ): n = len ( str ) - 1 ; # Start from rightmost position while (n > 0 ): # If the current character is ‘b’ or ‘d’, # change to ‘a’ or ‘c’ respectively and # break the loop if ( str [n] = = 'd' ): str = str [ 0 :n] + 'c' + str [n + 1 :]; break ; if ( str [n] = = 'b' ): str = str [ 0 :n] + 'a' + str [n + 1 :]; break ; # If the current character is ‘a’ or ‘c’, # change it to ‘b’ or ‘d’ respectively if ( str [n] = = 'a' ): str = str [ 0 :n] + 'b' + str [n + 1 :]; else if ( str [n] = = 'c' ): str = str [ 0 :n] + 'd' + str [n + 1 :]; n - = 1 ; return str ; # Driver Code if __name__ = = '__main__' : str = "aacbddc" ; print ( "Left of" , str , "is" , findLeft( str )); # This code is contributed by PrinciRaj1992 |
C#
using System; using System.Text; // C# program to return previous element // in an expanding matrix public class GFG { // Returns left of str in an expanding matrix // of a, b, c and d. public static StringBuilder findLeft(StringBuilder str) { int n = str.Length; // Start from rightmost position while (n > 0) { n--; // If the current character is b or d, // change to a or c respectively and // break the loop if (str[n] == 'd' ) { str[n] = 'c' ; break ; } if (str[n] == 'b' ) { str[n] = 'a' ; break ; } // If the current character is a or c, // change it to b or d respectively if (str[n] == 'a' ) { str[n] = 'b' ; } else if (str[n] == 'c' ) { str[n] = 'd' ; } } return str; } // driver program to test above method public static void Main( string [] args) { StringBuilder str = new StringBuilder( "aacbddc" ); Console.Write( "Left of " + str + " is " + findLeft(str)); } } // This code is contributed by Shrikant13 |
PHP
<?php // PHP program to return previous element in an expanding // matrix. // Returns left of str in an expanding matrix of // a, b, c and d. function findLeft( $str ) { $n = strlen ( $str ); // Start from rightmost position while ( $n --) { // If the current character is ‘b’ or ‘d’, // change to ‘a’ or ‘c’ respectively and // break the loop if ( $str [ $n ] == 'd' ) { $str [ $n ] = 'c' ; break ; } if ( $str [ $n ] == 'b' ) { $str [ $n ] = 'a' ; break ; } // If the current character is ‘a’ or ‘c’, // change it to ‘b’ or ‘d’ respectively if ( $str [ $n ] == 'a' ) $str [ $n ] = 'b' ; else if ( $str [ $n ] == 'c' ) $str [ $n ] = 'd' ; } return $str ; } // Driver Code $str = "aacbddc" ; echo "Left of " . $str . " is " . findLeft( $str ); return 0; ?> |
Javascript
<script> // Javascript program to return previous element // in an expanding matrix String.prototype.replaceAt = function (index, replacement) { return this .substr(0, index) + replacement + this .substr(index + replacement.length); } // Returns left of str in an expanding matrix // of a, b, c and d. function findLeft(str) { let n = str.length; // Start from rightmost position while (n > 0) { n--; // If the current character is b or d, // change to a or c respectively and // break the loop if (str[n] == 'd' ) { str = str.replaceAt(n, 'c' ); break ; } if (str[n] == 'b' ) { str = str.replaceAt(n, 'a' ); break ; } // If the current character is a or c, // change it to b or d respectively if (str[n] == 'a' ) str = str.replaceAt(n, 'b' ); else if (str[n] == 'c' ) str = str.replaceAt(n, 'd' ); } return str; } // driver program to test above method let str = "aacbddc" ; document.write( "Left of " + str + " is " + findLeft(str)); // This code is contributed by rag2127. </script> |
输出：
Left of aacbddc is aacbdcd
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