LeetCode — Reverse Integer (C++ Java Python)

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题目:http://oj.leetcode.com/problems/reverse-integer/ 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? 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). 题目翻译: 反转整数。 例1:X = 123,返回321 例2:X = -123,返回-321 你有没有想过这些呢? 这里有一些编码之前要问的好问题。如果你已经考虑了这些,奖励积分给你! 如果整数的最后一位是0,应该输出什么?例如10,100的情况。 你有没有注意到反转后的整数可能溢出?假设输入的是一个32位的整数,则1000000003反转后溢出。应该如何处理这种情况? 抛出一个异常?不错,但如果不允许抛出异常呢?那么你必须重新设计函数(即添加一个额外的参数)。 分析:         考虑负数的处理,逐位反转。 C++实现: class Solution { public: int reverse(int x) { // Note: The Solution object is instantiated only once and is reused by each test case. int result = 0; int flag = 0; if(x < 0) { flag = 1; x = -x; } int lastDigit = 0; while(x > 0) { lastDigit = x - x / 10 * 10; result = result * 10 + lastDigit; x /= 10; } if(flag == 1) { result = - result; } return result; } };Java实现: public class Solution { public int reverse(int x) { // Note: The Solution object is instantiated only once and is reused by each test case. int result = 0; int flag = 0; if (x < 0) { flag = 1; x = -x; } int lastDigit = 0; while (x > 0) { lastDigit = x - x / 10 * 10; result = result * 10 + lastDigit; x /= 10; } if (flag == 1) { result = -result; } return result; } }Python实现: class Solution: # @return an integer def reverse(self, x): result = 0 flag = 0 if x < 0: flag = 1 x = -x while x > 0: lastDigit = x - x / 10 * 10 result = result * 10 + lastDigit x /= 10 if flag == 1: result = -result return result        感谢阅读,欢迎评论!

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