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.
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| Example 1:
Input: low = 100, high = 300
Output: [123,234]
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| Example 2:
Input: low = 1000, high = 13000
Output: [1234,2345,3456,4567,5678,6789,12345]
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Constraints:
- 10 <= low <= high <= 10^9
Solution Sliding Window#
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| class Solution {
public List<Integer> sequentialDigits(int lo, int hi) {
int lenLo = getLen(lo);
int lenHi = getLen(hi);
int i = lenLo;
List<Integer> res = new ArrayList<>();
while (i <= lenHi) {
res.addAll(buildSeqNums(i, lo, hi));
i++;
}
return res;
}
int getLen(int n) {
return Integer.toString(n).length();
}
String SAMPLE = "123456789";
List<Integer> buildSeqNums(int n, int lo, int hi) {
if (n == 0) return Collections.emptyList();
int start = 0;
List<Integer> res = new ArrayList<>();
while (start + n < SAMPLE.length() + 1) {
int val = Integer.parseInt(SAMPLE.substring(start, start + n));
if (val >= lo && val <= hi) {
res.add(val);
}
start++;
}
return res;
}
}
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