LeetCode第7题:
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.
解题思路:
假如:x=1234,y=0
1)y乘以10,x把末位4移除掉,加给y(x=123,y=4)
2)y乘以10,x把末位3移除掉,加给y(x=12,y=43)
3)y乘以10,x把末位2移除掉,加给y(x=1,y=432)
4)y乘以10,x把末位1移除掉,加给y(x=0,y=4321)
代码:
int y = 0; while(x/10!=0){ y*=10; y+=x%10; x/=10; } y=y*10 +x;
结果报错
原因是int的范围是-2147483647~2147483648
leetcode给的input是1534236469,倒转之后是9646324351,导致溢出了。所以要加上溢出判断
最后的代码
class Solution { public int reverse(int x) { int negative = 1; if(x <= Integer.MIN_VALUE) return 0; if(x < 0){ x = -x; negative = -1; } long y = 0; while(x/10!=0){ y*=10; y+=x%10; x/=10; } y=y*10 +x; if(y > Integer.MAX_VALUE) return 0; else return (int)y * negative; } }
这也是很坑啊,溢出就输出0,鬼知道啊
原文地址:https://www.cnblogs.com/anni-qianqian/p/9057725.html
时间: 2024-10-08 00:15:55