1291. Sequential Digits (M)

Sequential Digits (M)

题目

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

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题意

在给定范围内找到所有整数，是这些整数满足每一位上的数字都比前一位的数字大1。

思路

所有这样的数字一共才不到一百个，直接遍历所有这样的整数，判断是否在指定范围里。

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代码实现

Java

排序

class Solution {
    public List<Integer> sequentialDigits(int low, int high) {
        List<Integer> list = new ArrayList<>();
        int first = 1, num = 1;
        while (first <= 8) {
            if (num >= low && num <= high) {
                list.add(num);
            }
            if (num > high || num % 10 == 9) {
                num = ++first;
                continue;
            }
            num = num * 10 + num % 10 + 1;
        }
        Collections.sort(list);
        return list;
    }
}

不排序

class Solution {
    public List<Integer> sequentialDigits(int low, int high) {
        List<Integer> list = new ArrayList<>();
        Queue<Integer> q = new LinkedList<>();
        for (int i = 1; i < 9; i++) {
            q.offer(i);
        }
        while (!q.isEmpty()) {
            int num = q.poll();
            if (num >= low && num <= high) {
                list.add(num);
            }
            if (num <= high && num % 10 < 9) {
                q.offer(num * 10 + num % 10 + 1);
            }
        }
        return list;
    }
}