Target number (with.Java)

I learned about the target number (with.Java).

I want to solve the coding test problem, find out how to solve it differently from the retrospective of the problem I solved, and get to know.

Let’s get to the problem first.

Problem

There are n nonnegative integers.

You want to create a target number by adding or subtracting these integers appropriately without changing the order.

For example, to create a number 3 with [1, 1, 1, 1, 1, 1], you can use the following five methods.

-1+1+1+1+1 = 3

+1-1+1+1+1 = 3

+1+1-1+1+1 = 3

+1+1+1-1+1 = 3

+1+1+1+1-1 = 3

Write a solution function to return the number of ways to create a target number by adding and subtracting numbers appropriately when the array numbers contain usable numbers, and target number targets are given as parameters.

Restrictions

• The number of numbers given is 2 or more and 20 or less.
• Each number is a natural number greater than or equal to 1 and less than or equal to 50.
• The target number is a natural number of 1 or more and 1000 or less.

Input/output Examples

problem solving

class Solution {
    static int answer;
    public int solution(int[] numbers, int target) {
        answer = 0;
        dfs(numbers, target, 0, 0);
        return answer;
    }
    public void dfs(int[] numbers, int target, int depth, int total){
        if(numbers.length == depth){
            if(total == target){
                answer++;
            }
        }else{
            dfs(numbers, target, depth + 1, total + numbers[depth]);
            dfs(numbers, target, depth + 1, total - numbers[depth]);
        }
    }
}

Solution Description

1. Declaring Classes and Variables class Solution: Class for troubleshooting. static intanswer: The static variable answer stores the number of times a target number can be created.
2. solution method public int solution (int[] numbers, int target): Returns the number of times a target number can be created by taking a given numeric array of numbers and a target numeric target as input. answer = 0: Initializes the number of cases. dfs(numbers, target, 0, 0): Initiates a depth-first search; sets the initial depth to 0, and the initial total to 0. return answer—Returns the number of cases finally calculated.
3. dfs method public void dfs (int[] numbers, int target, int depth, inttotal): a recursive method that performs depth-first navigation. if (number.length == depth): Verify that the end of the array has been reached. if (total == target): Increase the number of cases where the current total equals the target number target. else: If you have not reached the end of the array, proceed to the next step. dfs (number, target, depth + 1, total + numbers [depth]): Explore the case of adding the current number. dfs (number, target, depth + 1, total-number[depth]): Explore the case where the current number is subtracted.

Conclusion

This code uses DFS to explore all possible cases and to calculate the number of cases that match the target number.

Given a given numerical array and a target number, this code can efficiently determine the number of cases in which a target number can be created.