leetcode 7.Reverse Integer
2016-06-18 18:31
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//第一种方法比较简单也更快,第二种方法简直就是苦逼的方法
int reverse(int x){
long ret =0;
while(x){
ret = ret*10+x%10;
x = x/10;
}
if( (ret>=INT32_MIN) && (ret<=INT32_MAX) ){
return ret;
}else{
return0;
}
}
//一个test case , 即翻转过来是-2147483648似乎并没有测试到,所以通过了。。。
int reverse(int x){
int sign = x>0?1:-1;
x = x*sign;
char ret[16] = {'0','0','0','0','0','0','0','0','0','0'};
int len =9;
while(x){
ret[len--] = x%10+'0';
x = x/10;
}
ret[10] ='\0';
for(int i=len+1,j=9;i<j;i++,j--){
char tmp = ret[i];
ret[i] = ret[j];
ret[j] = tmp;
}
char maxx[16] = {'2','1','4','7','4','8','3','6','4','7'};
bool flag=false;
for(int i=0;i<10;i++){
if(ret[i]<maxx[i]){
flag = true;
break;
}
if(ret[i]>maxx[i]){
flag = false;
break;
}
if(ret[i]==maxx[i]){
continue;
}
}
if(flag ==false){
return0;
}else{
return sign*atoi(ret);
}
}
int reverse(int x){
long ret =0;
while(x){
ret = ret*10+x%10;
x = x/10;
}
if( (ret>=INT32_MIN) && (ret<=INT32_MAX) ){
return ret;
}else{
return0;
}
}
//一个test case , 即翻转过来是-2147483648似乎并没有测试到,所以通过了。。。
int reverse(int x){
int sign = x>0?1:-1;
x = x*sign;
char ret[16] = {'0','0','0','0','0','0','0','0','0','0'};
int len =9;
while(x){
ret[len--] = x%10+'0';
x = x/10;
}
ret[10] ='\0';
for(int i=len+1,j=9;i<j;i++,j--){
char tmp = ret[i];
ret[i] = ret[j];
ret[j] = tmp;
}
char maxx[16] = {'2','1','4','7','4','8','3','6','4','7'};
bool flag=false;
for(int i=0;i<10;i++){
if(ret[i]<maxx[i]){
flag = true;
break;
}
if(ret[i]>maxx[i]){
flag = false;
break;
}
if(ret[i]==maxx[i]){
continue;
}
}
if(flag ==false){
return0;
}else{
return sign*atoi(ret);
}
}
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