Reverse Integer

将一个整数中的数字进行颠倒，当颠倒后的整数溢出时，返回 0 (标记为 32 位整数)。

样例

给定 x = 123，返回 321
给定 x = -123，返回 -321

Solution

注意溢出即可

Complexity

时间复杂度 O(n)，空间复杂度 O(1)

Solution

public class Solution {
    public int reverse(int x) {
        int flag = 0;
        if (x > 0){
            flag = 1;
        } else if (x < 0) {
            flag = -1;
        }
        
        long tmp = (long)x;
        long answer = 0;
        tmp *= flag;
        while(tmp > 0){
            answer = answer * 10 + tmp % 10;
            tmp = (tmp - tmp % 10) / 10;
            if (answer > Integer.MAX_VALUE || flag * answer < Integer.MIN_VALUE) return 0;
        }
        return (int)answer*flag;
    }
}

Reference

class Solution:
    # @param {int} n the integer to be reversed
    # @return {int} the reversed integer
    def reverseInteger(self, n):
        # Write your code here
        res = 0
        if n >= 0:
            pos = True
        else:
            pos  = False
            n = -n
        while not n == 0:
            if res > 214748364:
                return 0
            else:
                res = res*10 + n%10
                n = n/10
        if pos:
            return res
        else:
            return -res