An algorithm for determining palindrome numbers with space complexity of O (1)

An algorithm for determining the palindrome number of O (1) In space complexity

To realize the spatial complexity of O (1) palindrome number determination, input is the integer constant, the output is required to determine whether the palindrome number.

The required format is as follows:

 Public boolean ispalindrome (int  x) {    //Your Judge code    }

Second, the concept

Palindrome number (palindrome) definition: set N is an arbitrary natural number. If the numbers of n are reversed and the natural number N1 is equal to N, then N is a palindrome number. For example, if n=1234321, then n is a palindrome number, but if n=1234567, then n is not a palindrome number.

　　Characteristics:

1. Negative numbers, decimals are not palindrome numbers;

2. Palindrome can be an odd number or even a number of numbers;

Third, analysis

According to the definition of palindrome number, we should pay attention to the following points in the design of the program in this subject:

1. When the number of palindrome inverted after the overflow may exist;

2. The spatial complexity of O (1) requires variable definition number can not be too many, if an input, then the custom variable should be defined as 1 and within;

Four, the algorithm

Set the input data to X.

1. If x is less than 0 or x is 10 N (n>=1), the return is false;

2. Set the comparison variable rev for 0,rev each increment of the last digit of x and X to remove the last digit each time, if Rev is less than or equal to X, then continue 2 steps, otherwise enter step 3;

3. Output The final result, if Rev equals X or Rev except 10 rounding equals X, then x is a palindrome number, otherwise it is not a palindrome number.

V. Java code

 Public classPalindromenumber { Public Static voidMain (string[] args) {Palindromenumber P=NewPalindromenumber (); System.out.println (P.ispalindrome (32233223)); }         Public BooleanIspalindrome (intx) {if(X < 0 | | (x! = 0 && x% 10 = = 0)) {            return false; }        intRev = 0;  while(X >rev) {Rev= Rev * + x% 10; X= X/10; }        return(x = Rev | | x = = REV/10); }}

Six, advantages

1. The algorithm defines only an additional temporary variable rev, without defining a string and then comparing it by loop;

2. Through the continuous growth of rev and the shortening of X, the two end up to the length of X/2, no overflow situation;

3. The performance is very high, the occupied time space is very low.

An algorithm for determining palindrome numbers with space complexity of O (1)