LeetCode:Reverse Integer

Posted on 2015-10-09 In 编程 Disqus:

Total Accepted: 100793 Total Submissions: 429248 Difficulty: Easy
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. Update (2014-11-10): Test cases had been added to test the overflow behavior.

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Math

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(E) String to Integer (atoi)

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来源： https://leetcode.com/problems/reverse-integer/

代码(Accepted):

class Solution {
public:
    int reverse(int x) {
        int ret = 0;
        while(x!=0){
            if(ret>INT_MAX/10||ret<INT_MIN/10){
                return 0;
            }
            ret = ret*10 + x%10;
            x = x/10;
        }
        return ret;
    }
};

我感觉最准确的应该是下面这个

class Solution {
public:
    int reverse(int x) {
        int ret = 0;
        while(x!=0){
            if(ret>INT_MAX/10-x%10/10||ret<INT_MIN/10-x%10/10){
                return 0;
            }
            ret = ret*10 + x%10;
            x = x/10;
        }
        return ret;
    }
};

Solution:

To check for overflow/underflow, we could check if ret > 214748364 or ret < –214748364 before multiplying by 10. On the other hand, we do not need to check if ret == 214748364, why?

本题解析：

本题一定要把判断加在while循环之中，并且要在乘10之前判断是否超出范围。

遗留问题：

1.两段代码更精确的感觉应该是下面那个。
2.运行时间8ms，比较长，有待优化。(your runtime beats 24.74% of cpp submissions.)

LeetCode:Reverse Integer

代码(Accepted):

Solution:

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本题解析：

遗留问题：

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