int reverse(int x) {
int x_re = 0;
while(abs(x/10) > 0)
{
if(isOverflowInt(x_re, x%10))
return 0;
x_re = x_re*10 + x%10;
x /= 10;
}
if(isOverflowInt(x_re, x))
return 0;
x_re = x_re *10 +x;
return x_re;
}
//判断是否溢出的函数:溢出位true,未溢出为false
bool isOverflowInt(int x, int pop)
{
bool is=false;
if(x>0){
is = INT_MAX/10 < x ? true : false;
if((!is) && (INT_MAX/10 == x))
is = pop > 7 ? true : false;
}
if(x <0){
is = INT_MIN/10 > x ? true : false;
if((!is) && (INT_MIN/10 == x))
is = pop < -8 ? true : false;
}
return is;
}
leetcode-32位整数反转并检查溢出
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转载自blog.csdn.net/mixiaoxinmiss/article/details/80969895
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