def reverse(x: int) -> int: # 设置32位整数的上下限 INT_MAX, INT_MIN = 2**31 - 1, -2**31 # 记录符号,正数为1,负数为-1 sign = 1 if x > 0 else -1 x *= sign # 去掉负号进行翻转处理:把正数和负数都当正数来对待 # 翻转数字 reversed_x = 0 while x: reversed_x = reversed_...
题意:翻转一个int数,注意两点,一是越界,而是0的问题 classSolution {publicintreverse(intx) {if(x == -2147483648)return0;booleanflag =false;if(x < 0) flag=true; x=Math.abs(x);longres = 0;while(x != 0) { res= res*10 + x%10; x/= 10;if(res >Integer.MAX_VALUE) { res= 0;...
classSolution {public:intreverse(intx) {if(x<=9&& x>=-9)returnx;intMAX = INT_MAX/10;intMIN = INT_MIN/10;intresult =0;while(x !=0){if( result>MAX || result<MIN||(x >0&& INT_MAX-x%10<result*10)|| (x<0&& INT_MIN-x%10>result*10) )return0; result= result*10+ x%...
class Solution: # @return an integer def reverse(self, x): revx = int(str(abs(x))[::-1]) if revx > math.pow(2, 31): return 0 else: return revx * cmp(x, 0) 1. 2. 3. 4. 5. 6. 7. 8.
## 解法一:转换为字符串,要考虑溢出 ## 特殊的案例:< 2**31-1 class Solution: def reverse(self, x: int) -> int: if x>=0: ans = int(str(x)[::-1]) else: ans = -int( str(abs(x))[::-1] ) ## 不用用 -x? ## 考虑溢出 return ans if ans<2**31-1 and ans>=-2**31...
Reverse Integer Leetcode 7. Reverse Integer 题目说明 代码部分1 代码部分2 题目说明 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: 1......
class Solution { /** * @param Integer $x * @return Integer */ function reverse($x) { $max = pow(2,31)-1;//最大 $min = pow(-2,31);//最小 if($x>=0){ $num = (Integer)strrev($x);//反转 }else{ $num = '-'.(Integer)strrev(trim($x,'-'));//反转 } return $num>...
class Solution { public: int reverse(int x) { if (x == -2147483648) return 0; int ax = x; if (x < 0) ax = -x; //if (ax > numeric_limits<int>::max()) return 0; int y = r2(ax); if (x < 0) return -y;
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows. class Solution(object): def reverse(self, x): if x == 1534236469 or abs(x) == 1563847412: return 0 else:
给你一个 32 位的有符号整数x,返回将x中的数字部分反转后的结果。 如果反转后整数超过 32 位的有符号整数的范围[−231, 231− 1],就返回 0。 假设环境不允许存储 64 位整数(有符号或无符号)。 示例1: 输入:x = 123输出:321 示例2: 输入:x = -123输出:-321 ...