Given an integer, return its base 7 string representation.
Example 1:
Input: 100 Output: "202"
Example 2:
Input: -7 Output: "-10"
Note: The input will be in range of [-1e7, 1e7].
这道题给了我们一个数,让我们转为七进制的数,而且这个可正可负。那么我们想如果给一个十进制的100,怎么转为七进制。我会先用100除以49,商2余2。在除以7,商0余2,于是就得到七进制的202。其实我们还可以反过来算,先用100除以7,商14余2,然后用14除以7,商2余0,再用2除以7,商0余2,这样也可以得到202。这种方法更适合于代码实现,要注意的是,我们要处理好负数的情况,参见代码如下:
解法一:
class Solution { public: string convertToBase7(int num) { if (num == 0) return "0"; string res = ""; bool positive = num > 0; while (num != 0) { res = to_string(abs(num % 7)) + res; num /= 7; } return positive ? res : "-" + res; } };
上面的思路也可以写成迭代方式,非常的简洁,仅要三行就搞定了,参见代码如下:
解法二:
class Solution { public: string convertToBase7(int num) { if (num < 0) return "-" + convertToBase7(-num); if (num < 7) return to_string(num); return convertToBase7(num / 7) + to_string(num % 7); } };
参考资料:
https://discuss.leetcode.com/topic/78934/1-line
https://discuss.leetcode.com/topic/78972/simple-java-oneliner-ruby
https://discuss.leetcode.com/topic/78935/java-1-liner-standard-solution