LeetCode 145 Reverse Integer
2014-11-01 16:20
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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).
分析:
这题是 Palindrome Number 的基础,即通过求商和求余来翻转整数。
public class Solution {
public int reverse(int x) {
if(Math.abs(x) <10)
return x;
boolean pos;
if(x>0)
pos=true;
else
pos=false;
x=Math.abs(x);
int res=0;
while(x!=0){
int digit=x%10;
x = x/10;
res=res*10+digit;
}
return (pos==true)? res : (-res);
}
}
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).
分析:
这题是 Palindrome Number 的基础,即通过求商和求余来翻转整数。
public class Solution {
public int reverse(int x) {
if(Math.abs(x) <10)
return x;
boolean pos;
if(x>0)
pos=true;
else
pos=false;
x=Math.abs(x);
int res=0;
while(x!=0){
int digit=x%10;
x = x/10;
res=res*10+digit;
}
return (pos==true)? res : (-res);
}
}
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