Description:
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.
Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
算法思想：
循環從低位到高位取得每個數的值，在每次循環時，讓該數乘10累加，最終可以得到與原數逆序的數
分別處理負數情況、逆序後數字越界情況
C++ Code:
#include <iostream>
#include "stdlib.h"
#include "math.h"
using namespace std;
class Solution {
public:
int reverse(int x) {
if(x==-2147483648){ //最小負數，由於轉換爲正數時會越界，單獨處理
return 0;
}
int maxNum[10]={2,1,4,7,4,8,3,6,4,7};//最大正數
bool flag=false;
bool isNeg=false;
if(x<0){
x=-x;
isNeg=true;
}
if(x/1000000000!=0){//判斷是否達到最大數位數
flag=true;
}
int s=0,j=0,m;
while(x!=0){
m=x%10;
if(flag){
if(m>maxNum[j]){//轉置後越界，返回0
return 0;
}else if(m<maxNum[j]){
flag=false;
}
j++;
}
x=x/10;
s=s*10+m;
}
if(isNeg){
return -s;
}
return s;
}
};
int main(){
int s=-2147483648;
Solution sol;
int res=sol.reverse(s);
cout<<res<<endl;
}