【數論】B055_LC_順次數（構造法）

一、Problem

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.

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

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

Constraints:

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

二、Solution

方法一：構造

從 1 開始，不斷枚舉，構造，比較簡單…

class Solution {
	boolean ck(int num) {
		String s = num + "";
		for (int i = 1; i < s.length(); i++) {
			if (s.charAt(i-1) >= s.charAt(i))
				return false;
		}
		return true;
	}
    public List<Integer> sequentialDigits(int low, int hig) {
    	String lo = low + "", hi = hig + "";
    	int min = Integer.parseInt(lo), max = Integer.parseInt(hi);
    	List<Integer> res = new LinkedList<>();
        
    	for (int i = 1; i < 10; i++) {
    		int num = i, j = i + 1;
    		while (true) {
	    		num = num * 10 + j;
	    		if (min <= num && num <= max && ck(num)) {
	    			res.add(num);
	    		}
	    		if (num > max)
	    			break;
	    		j++;
    		}
    	}
        Collections.sort(res);
    	return res;
    }
}

因爲構造出來的數字就是 “有序的”，所以不需要再檢查是否是 順次的…

class Solution {
    public List<Integer> sequentialDigits(int low, int hig) {
    	List<Integer> res = new LinkedList<>();
    	for (int i = 1; i < 10; i++) {
    		int num = i;
    		for (int j = i + 1; j < 10; j++) {
	    		num = num * 10 + j;
	    		if (low <= num && num <= hig)
	    			res.add(num);
	    		if (num > hig)
	    			break;
    		}
    	}
        Collections.sort(res);
    	return res;
    }
}

複雜度分析

• 時間複雜度：$O(1)$，
• 空間複雜度：$O(1)$，