LeetCode - 7 Reverse Integer

题目:

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

Have you thought about this?

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?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

解答:

1.此题的代码相对简单, 但是要考虑overflow的问题

- (1) 运用long可以cast int的特性, 并且范围也比int大来解决overflow问题.

- (2) 运用overflow之后/10的结果与原来的不相同来解决问题

 

代码: - (1)

public class solution {

　　public int reverse(int x) { // x is the inquiry

　　　　long reversed = 0;

 

　　　　while (x != 0) {
　　　　　　reversed = x%10 + reversed * 10;

　　　　　　x = x /10;

　　　　}

 

　　　　if (reversed > Integer.MAX_VALUE || reversed < Integer.MIN_VALUE) {

　　　　　　return 0; //此处需要问面试官需要什么样的return, 可以是0, 可以是-1

　　　　}

 

　　　　return reversed;

　　}

}

 

代码: - (2)

public class Solution {
　　public int reverse(int x) {
　　　　int reversed = 0;
　　　　while (x != 0) {
　　　　　　int test = x % 10 + reversed * 10;
　　　　　　x = x / 10;
　　　　　　if (test/10 != reversed) {
　　　　　　　　reversed = 0;
　　　　　　　　break;
　　　　　　} // 此处处理overflow问题
　　　　　　reversed = test;
　　　　}
　　　　return reversed;
　　}
}