【LeetCode】- Reverse Integer(将一个整数反转)
2014-01-05 22:11
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[ 问题: ]
Reverse digits of an integer.
Example1: x
= 123, return 321
Example2: x = -123, return -321
题解:给定一个整数,将这个整数的数字旋转位置。
[ 分析 : ]
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).
注意点1:如果一个整数最后一位位0,比如:10100,那么反转后将变成101。
注意点2:如果输入的数为1000000003,那么反转后将会出现溢出,需要注意。
[ 解法: ]
Reverse digits of an integer.
Example1: x
= 123, return 321
Example2: x = -123, return -321
题解:给定一个整数,将这个整数的数字旋转位置。
[ 分析 : ]
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).
注意点1:如果一个整数最后一位位0,比如:10100,那么反转后将变成101。
注意点2:如果输入的数为1000000003,那么反转后将会出现溢出,需要注意。
[ 解法: ]
/* * 假设输入123: * 123 -> 3 * 12 -> 30 + 2 * 1 -> 320 + 1 */ public class Solution { public int reverse(int x) { int result = 0; while (x != 0) { result = result * 10 + x % 10; // 每一次都在原来结果的基础上变大10倍,再加上余数 x = x / 10; // 对x不停除10 } return result; } public static void main(String[] args) { Scanner scanner = new Scanner(new BufferedInputStream(System.in)); while (scanner.hasNext()) { int num = scanner.nextInt(); System.out.println(new Solution().reverse(num)); } } }
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