Leetcode在线编程 reverse-integer

题目链接

reverse-integer

题目描述

Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
click to show spoilers.
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).

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题意

将给定的数翻转，题意其实很简单，但需要注意在翻转的时候会有前置0或者越界，越界的话返回0就可以了

解题思路

使用字符流stringstream去解决数据转换问题，减少了代码量，每次使用stringstream都需要去clear()
将整数x转变为字符串
int x
stringstream ss ;
ss << x;
string s;
ss >> s;

AC代码

class Solution {public:    int reverse(int x) {        stringstream ss ;        ss << x;        string s;        ss >> s;        long long f[2];        long  long a = ((long long)(1) <<32) - 1;        long long  b =  ((long long)1 << 32);        f[0] = a;        f[1] = b;        int len = s.length();        int flag = 0;        if(s[0]=='-')            flag = 1;        for(int i = flag,j = 0;i< len/2 + flag;i++,j++)            swap(s[i],s[len-1-j]);        string tmp1;        tmp1 = s.substr(flag,len);        string tmp2;        ss.clear();        ss << f[flag];         ss >> tmp2;        if(tmp2.length()<tmp1.length()||tmp1.length()==tmp2.length()&&tmp2<tmp1)           return 0;        ss.clear();        ss << s;        ss >> x;        return x;    }};