【大数库暴力法】7. Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321
Example 2:

Input: -123
Output: -321
Example 3:

Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

class Solution {
public int reverse(int x) {
if(x == 0) return 0;
else {
StringBuilder str = new StringBuilder(String.valueOf(x));
StringBuilder builder = str.reverse();
builder = x < 0 ? (builder.deleteCharAt(builder.length()-1)) : builder;
boolean flag = true;
for(int i = 0; builder.length() > 0; ) {
if(flag) {
if(builder.charAt(i) != '0') {
flag = false; break;
}
if(builder.charAt(i) == '0')builder.deleteCharAt(i);
}
}
long rest = Long.valueOf(String.valueOf(builder));
rest = x < 0 ? -rest : rest;
return (rest < Integer.MIN_VALUE || rest > Integer.MAX_VALUE) ? 0 : (int)rest;
}
}
}

用数学方法的话时间更快些吧。

下面是leetcode上用户Jinx_boom写的,可以看看。

public int reverse(int x) {
int res = 0;
boolean flag = x > 0;
while(x!=0){
if(Math.abs(res)>214748364||Math.abs(res)==214748364&&Math.abs(x%10)>((flag)?7:8))   return 0;
res = res*10+x%10;
x/=10;
}
return res;
}