LeetCode 2094. Finding 3-Digit Even Numbers

原题链接在这里：https://leetcode.com/problems/finding-3-digit-even-numbers/

题目：

You are given an integer array digits, where each element is a digit. The array may contain duplicates.

You need to find all the unique integers that follow the given requirements:

• The integer consists of the concatenation of three elements from digits in any arbitrary order.
• The integer does not have leading zeros.
• The integer is even.

For example, if the given digits were [1, 2, 3], integers 132 and 312 follow the requirements.

Return a sorted array of the unique integers.

Example 1:

Input: digits = [2,1,3,0]
Output: [102,120,130,132,210,230,302,310,312,320]
Explanation: All the possible integers that follow the requirements are in the output array. 
Notice that there are no odd integers or integers with leading zeros.

Example 2:

Input: digits = [2,2,8,8,2]
Output: [222,228,282,288,822,828,882]
Explanation: The same digit can be used as many times as it appears in digits. 
In this example, the digit 8 is used twice each time in 288, 828, and 882. 

Example 3:

Input: digits = [3,7,5]
Output: []
Explanation: No even integers can be formed using the given digits.

Constraints:

• 3 <= digits.length <= 100
• 0 <= digits[i] <= 9

题解：

From 100 to 999, for each even number i, check its digits all appear in the map.

The map is precalculated by digits array.

Time Complexity: O(n). n = 900/2.

Space: O(1).

AC Java:

class Solution {
    public int[] findEvenNumbers(int[] digits) {
        int [] count = new int[10];
        for(int d : digits){
            count[d]++;
        }
        
        List<Integer> list = new ArrayList<>();
        for(int i = 100; i < 1000; i+=2){
            int a = i % 10; 
            int b = (i / 10) % 10;
            int c = i / 100;
            
            count[a]--;
            count[b]--;
            count[c]--;
            if(count[a] >=0 && count[b] >= 0 && count[c] >= 0){
                list.add(i);
            }
            
            count[a]++;
            count[b]++;
            count[c]++;
        }
        
        int [] res = new int[list.size()];
        for(int i = 0; i < res.length; i++){
            res[i] = list.get(i);
        }
        
        return res;
    }
}