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; }