[Swift]LeetCode1215. 步进数 | Stepping Numbers
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A Stepping Number is an integer such that all of its adjacent digits have an absolute difference of exactly 1. For example, 321 is a Stepping Number while 421 is not.
Given two integers low and high, find and return a sorted list of all the Stepping Numbers in the range [low, high] inclusive.
Example 1:
Input: low = 0, high = 21 Output: [0,1,2,3,4,5,6,7,8,9,10,12,21]
Constraints:
0 <= low <= high <= 2 * 10^9
如果一个整数上的每一位数字与其相邻位上的数字的绝对差都是 1，那么这个数就是一个「步进数」。
例如，321 是一个步进数，而 421 不是。
给你两个整数，low 和 high，请你找出在 [low, high] 范围内的所有步进数，并返回 排序后 的结果。
示例：
输入：low = 0, high = 21 输出：[0,1,2,3,4,5,6,7,8,9,10,12,21]
提示：
0 <= low <= high <= 2 * 10^9
Runtime: 152 ms
Memory Usage: 22.3 MB
1 class Solution {
2 var L:Int = 0
3 var S:[Int] = [Int]()
4
5 func countSteppingNumbers(_ low: Int, _ high: Int) -> [Int] {
6 self.L = high
7 var ans:[Int] = [Int]()
8 for i in 1...9
9 {
10 search(i)
11 }
12 if 0 >= low && 0 <= high
13 {
14 ans.append(0)
15 }
16 for i in S
17 {
18 if i >= low && i <= high
19 {
20 ans.append(i)
21 }
22 }
23 ans.sort()
24 return ans
25 }
26
27 func search(_ x:Int)
28 {
29 if x > L{return}
30 S.append(x)
31 let last:Int = x % 10
32 if last != 0
33 {
34 search(x * 10 + last - 1)
35 }
36 if last != 9
37 {
38 search(x * 10 + last + 1)
39 }
40 }
41 }