|
| 1 | +/* |
| 2 | +习题: |
| 3 | +Given a 32-bit signed integer, reverse digits of an integer. |
| 4 | + |
| 5 | +Example 1: |
| 6 | + |
| 7 | +Input: 123 |
| 8 | +Output: 321 |
| 9 | +Example 2: |
| 10 | + |
| 11 | +Input: -123 |
| 12 | +Output: -321 |
| 13 | +Example 3: |
| 14 | + |
| 15 | +Input: 120 |
| 16 | +Output: 21 |
| 17 | +Note: |
| 18 | +Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. |
| 19 | +For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows. |
| 20 | + |
| 21 | +*/ |
| 22 | +/* |
| 23 | +解析: |
| 24 | + |
| 25 | +这题其实理解起来不难,题目要求逆序输出。 |
| 26 | +逆序输出不难,但是难点是32bit ,一旦溢出要返回0。那怎么判断溢出? |
| 27 | + |
| 28 | +因为是32位的有符号的整数,最大的十进制就是-(2^31)-1 ~ 2^31 -1 |
| 29 | + |
| 30 | + |
| 31 | + |
| 32 | +*/ |
| 33 | + |
| 34 | +/* |
| 35 | +分解数位 |
| 36 | + |
| 37 | +%:分解成个位数字 |
| 38 | +/: 去除个位数字。 |
| 39 | +逆序数字就是:(分解个位数字之后)*10+个位数字 |
| 40 | + |
| 41 | +具体代码如下: |
| 42 | + |
| 43 | +*/ |
| 44 | + |
| 45 | +class Solution { |
| 46 | + public int reverse(int x) { |
| 47 | + int digit = 0; |
| 48 | + int result = 0; |
| 49 | + int sybol=0;//带符号,0代表负号,1代表正号。 |
| 50 | + |
| 51 | + //因为是x是带符号的,所以分情况讨论 |
| 52 | + if(x>0) { |
| 53 | + sybol=1; |
| 54 | + } |
| 55 | + else { |
| 56 | + sybol=0; |
| 57 | + x = -x; |
| 58 | + } |
| 59 | + |
| 60 | + //对于溢出的问题进行处理 |
| 61 | + if(x>=2147483647) { |
| 62 | + return 0; |
| 63 | + } |
| 64 | + |
| 65 | + while(x>0){ |
| 66 | + digit = x % 10; |
| 67 | + //对于溢出的问题进行处理。 |
| 68 | + if(result >2147483647/10.0) |
| 69 | + return 0; |
| 70 | + result = result*10+digit; |
| 71 | + x = x/10; |
| 72 | + } |
| 73 | + |
| 74 | + return (sybol==1)?result:-result; |
| 75 | + } |
| 76 | +} |
| 77 | + |
0 commit comments