编写一个程序来显示PHP中任意数字的倒数？
编写一个程序来反转整数的数字。
例子 ：
Input : num = 12345
Output: 54321
Input : num = 876
Output: 678可以使用迭代方式或递归方式来实现
迭代方法：
算法：
Input: num
(1) Initialize rev_num = 0
(2) Loop while num > 0
(a) Multiply rev_num by 10 and add remainder
of num divide by 10 to rev_num
rev_num = rev_num*10 + num%10;
(b) Divide num by 10
(3) Return rev_num例子：
num = 456213
rev_num = 0
rev_num = rev_num *10 + num%10 = 3
num = num/10 = 45621
rev_num = rev_num *10 + num%10 = 20 + 6 = 31
num = num/10 = 4562
rev_num = rev_num *10 + num%10 = 260 + 5 = 312
num = num/10 = 456
rev_num = rev_num *10 + num%10 = 2650 + 4 = 3126
num = num/10 = 45
rev_num = rev_num *10 + num%10 = 2650 + 4 = 31265
num = num/10 = 4
rev_num = rev_num *10 + num%10 = 2650 + 4 = 312654
num = num/10 = 0
程序：
PHP
1) {
$rev_num = $rev_num * 10 + $num % 10;
$num = (int)$num / 10;
}
return $rev_num;
}
// Driver Code
$num = 456213;
echo "Original number is :".$num;
echo "\r\n";
echo "Reverse of no. is ", reversDigits($num);
?>PHP
0) {
reversDigits((int)($num / 10));
$rev_num += ($num % 10) *
$base_pos;
$base_pos *= 10;
}
return $rev_num;
}
// Driver Code
$num = 456213;
echo "Original number is :".$num;
echo "\r\n";
echo "Reverse of no. is ",
reversDigits($num);
?>输出
Original number is :456213
Reverse of no. is 312654时间复杂度： O(log(n))，其中 n 是输入数。
辅助空间： O(1)
递归方法：
PHP
0) {
reversDigits((int)($num / 10));
$rev_num += ($num % 10) *
$base_pos;
$base_pos *= 10;
}
return $rev_num;
}
// Driver Code
$num = 456213;
echo "Original number is :".$num;
echo "\r\n";
echo "Reverse of no. is ",
reversDigits($num);
?>
输出
Original number is :456213
Reverse of no. is 312654时间复杂度： O(log(n))，其中 n 是输入数。