题目如下:

An integer has sequential digits if and only if each digit in the number is one more than the previous digit.

Return a sorted list of all the integers in the range [low, high] inclusive that have sequential digits. 

Example 1:

Input: low = 100, high = 300
Output: [123,234]

Example 2:

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

Constraints:

  • 10 <= low <= high <= 10^9

解题思路:和 【leetcode】1215.Stepping Numbers 思路是一样的,利用BFS的思想,把所有符合条件的数字找出来。

代码如下:

class Solution(object):
    def sequentialDigits(self, low, high):
        """
        :type low: int
        :type high: int
        :rtype: List[int]
        """
        res = []
        queue = range(1,10)
        while len(queue) > 0:
            val = queue.pop(0)
            if val >= low and val <= high:
                res.append(val)
            elif val > high:
                continue
            last = int(str(val)[-1])
            if last == 9:continue
            new_val = str(val) + str(last + 1)
            queue.append(int(new_val))
        return sorted(res)

 

相关文章:

  • 2022-02-11
  • 2021-05-29
  • 2021-11-07
  • 2021-10-20
  • 2022-01-28
  • 2021-10-22
  • 2022-01-13
  • 2021-07-25
猜你喜欢
  • 2021-12-28
  • 2022-12-23
  • 2021-07-07
  • 2021-11-23
  • 2021-12-29
  • 2021-12-27
相关资源
相似解决方案