2020 Ali practice written test - playing cards -DFS

2020 Ali practice written test - playing cards -DFS

Title Description

Today Xiaoqiang to come from a deck of cards in a stack, including A, 2,3, ..., 10 each of the four in which A represents 1. He pulled out some of the cards in a stack to Xiao Ming From this and tell each Xiaoming Some cards can be played in the following way:

• Single cards: a card, such as 3
• Pair: two cards of the same figure, for example, 77
• Straight: Five cards of consecutive numbers, such as A2345
• Three pairs: three consecutive pairs: for example, 334,455
  now Xiaoqiang want to know how many times you can play cards at least polish their hand.

Enter a description:

Line is a space-separated integers A _{. 1} , A ₂ , ... A ₁₀ , representing the card is A, 2, ..., the number 10.
0 <= A _{. 1} , A ₂ , ..., A ₁₀ <=. 4
guarantee hand At least one card

Output Description:

Only one per line integer answer

Example:

Input:

1 1 1 2 2 2 2 2 1 1

Export

3

Description:

Hit three straight, are: A2345,45678,678910

Depth-first Code:

public class Main {
    public static int min = Integer.MAX_VALUE;
    public static boolean check(int[] nums) {
        boolean flag = false;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] > 0) {
                flag = true;
                break;
            }
        }
        return flag;
    }
    public static void dfs(int[] nums, int count) {
    	//剪枝
    	if(count>min)
    	{
    		return;
    	}
        if (check(nums)) {
            int i = 0;
            while (nums[i] == 0) {
                i++;
            }
            //单张
            if (nums[i] > 0) {
                nums[i]--;
                dfs(nums, count+1);
                nums[i]++;
            }
            //对子
            if (nums[i] > 1) {
                nums[i] -= 2;
                dfs(nums,count+1);
                nums[i] += 2;
            }
            //五连
            if (i + 5 <= nums.length && nums[i] > 0 && nums[i + 1] > 0 && nums[i + 2] > 0 && nums[i + 3] > 0 && nums[i + 4] > 0) {
                for (int j = 0; j < 5; j++) {
                    nums[i + j]--;
                }
                dfs(nums, count+1);
                for (int j = 0; j < 5; j++) {
                    nums[i + j]++;
                }
            }
            //三双
            if (i + 3 <= nums.length && nums[i] > 1 && nums[i + 1] > 1 && nums[i + 2] > 1) {
                nums[i] -= 2;
                nums[i + 1] -= 2;
                nums[i + 2] -= 2;
                dfs(nums, count+1);
                nums[i] += 2;
                nums[i + 1] += 2;
                nums[i + 2] += 2;
            }
        } else {
            if(count<min)
            {
                min = count;
            }
        }
    }
    public static void main(String[] args) {
        int[] nums = new int[]{1, 1, 1, 2, 2, 2, 2, 2, 1, 1};
        dfs(nums, 0);
        System.out.println(min);
    }
}