done Beaconfire中国小哥中规中矩screening

https://drive.google.com/drive/folders/1BTKV760sLnrdtP8WhGJQaqpZjoaEZhUd?usp=sharing

如何避免kafka中的一个消息被消耗两次 -分区

hashmap内部的结构是什么样的，两个值相同的object当作key size会是多少

回溯法初始的index = 0，才能算出重复值 eg (1,3) (3,1)

复杂度：数组元素个数n的（target）次方

 

/**
Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.


Input: nums = [1,2,3], target = 4
Output: 7
Explanation:
The possible combination ways are:

(1, 1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(1, 3)
(2, 1, 1)
(2, 2)
(3, 1)
Note that different sequences are counted as different combinations.
result = [[1, 1, 1, 1], [1, 1, 2], [1, 3], [2, 2]]
*/
import java.util.*;

public class MyClass {
    public static void main(String args[]) {
      int[] nums = {1,2,3};
      int target = 4;
      
      List<List<Integer>> result = new ArrayList<List<Integer>>();
      //sort
      Arrays.sort(nums);
      
      backtrace(nums, 0, new ArrayList<>(), result, 0, target);
      
      System.out.println("result = " + result);
      System.out.println("result.size() = " + result.size());
    }
    
    public static void backtrace(int[] nums, int start, List<Integer> temp, 
    List<List<Integer>> result, int currSum, int target) {
        //exit case
        if (currSum == target) {
            result.add(new ArrayList<>(temp));
        }else if (currSum > target) {
            return ;
        }else {
            for (int i = 0; i < nums.length; i++) {
                //handle duplicate
                // if (temp.contains(nums[i]))
                //     continue;
                
                //backtrace
                temp.add(nums[i]);
                backtrace(nums, i, temp, result, currSum + nums[i], target);
                temp.remove(temp.size() - 1);
            }
        }
    }
}