【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;
    }