Leetcode: 【每日一题】- 2019-12-24 1291. 顺次数

[azl397985856]

我们定义「顺次数」为：每一位上的数字都比前一位上的数字大 1 的整数。

请你返回由 [low, high] 范围内所有顺次数组成的 有序 列表（从小到大排序）。

 

示例 1：

输出：low = 100, high = 300
输出：[123,234]
示例 2：

输出：low = 1000, high = 13000
输出：[1234,2345,3456,4567,5678,6789,12345]
 

提示：

10 <= low <= high <= 10^9

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/sequential-digits
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

[azl397985856]

Easy to understand Python Solution with slide window

1. computed all possible sequentialDigits, put into a list ins (no more than 64)
2. iterate ins, put satisfied number which is x >= low and x <= high into returned list

class Solution:
    def sequentialDigits(self, low: int, high: int) -> List[int]:
        numbers = "123456789"
        ins = []
        n = len(numbers)
        for length in range(1, n):
            for i in range(n - length):
                ins.append(int(numbers[i:i + length + 1]))
        return [x for x in ins if x >= low and x <= high]

[unclegem]

Java题解：打表

    public List<Integer> sequentialDigits(int low, int high) {

        int[] nums = new int[]{12, 23, 34, 45, 56, 67, 78, 89,
                        123, 234, 345, 456, 567, 678, 789,
                        1234, 2345, 3456,4567, 5678, 6789,
                        12345, 23456, 34567, 45678, 56789,
                        123456, 234567, 345678, 456789,
                        1234567, 2345678, 3456789,
                        12345678, 23456789,
                        123456789};
        List<Integer> res = new ArrayList<>();
        for (int num : nums)
            if (num >= low && num <= high)
                res.add(num);

        return res;
    }

[stale[bot]]

This issue has been automatically marked as stale because it has not had recent activity. It will be closed if no further activity occurs. Thank you for your contributions.