LeetCode: Reverse Integer
2016-10-03 23:53
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题目:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
这道题很简单,从低位向高位取整数每位上的数字,并把初始化为0的result值乘10加上这个取得的数字,时间复杂度为o(logn)。但是这里要特别注意整数范围的问题,因为一个整数的倒序很可能会溢出。代码:
Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
这道题很简单,从低位向高位取整数每位上的数字,并把初始化为0的result值乘10加上这个取得的数字,时间复杂度为o(logn)。但是这里要特别注意整数范围的问题,因为一个整数的倒序很可能会溢出。代码:
class Solution { public: int reverse(int x) { const int max=0x7fffffff; //int最大值 const int min=0x80000000; //int最小值 //不能直接声明为int类型,否则溢出时会出错 long long result=0; while(x!=0) { int temp=x%10; result=result*10+temp; if(result>max||result<min) //溢出 { if(result>0) return max; else return min; } x=x/10; } return result; } };
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