티스토리 뷰

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소스 코드는 여기 있습니다.
문제는 여기 있습니다.

Problem

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

The test cases are generated so that the answer can fit in a 32-bit integer.

Example 1:

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.

Example 2:

Input: nums = [9], target = 3
Output: 0

Constraints:

  • 1 <= nums.length <= 200
  • 1 <= nums[i] <= 1000
  • All the elements of nums are unique.
  • 1 <= target <= 1000

Follow up: What if negative numbers are allowed in the given array? How does it change the problem? What limitation we need to add to the question to allow negative numbers?

Solution

중복되지 않은 정수 배열과 타겟 정수가 주어질 때 더해서 타겟 정수가 되는 모든 조합의 개수를 구하는 문제입니다.

public class Solution {

  public int combinationSum4(int[] nums, int target) {
    if (nums == null || nums.length == 0) {
      return 0;
    }
    int[] dp = new int[target + 1];
    dp[0] = 1;
    for (int i = 0; i < dp.length; i++) {
      for (int num : nums) {
        if (i + num <= target) {
          dp[i + num] += dp[i];
        }
      }
    }
    return dp[target];
  }
}

dp를 이용해 간단히 풀이할 수 있습니다.

Test

package io.lcalmsky.leetcode.combination_sum_iv;

import static org.junit.jupiter.api.Assertions.assertAll;
import static org.junit.jupiter.api.Assertions.assertEquals;

import org.junit.jupiter.api.Test;

class SolutionTest {

  @Test
  public void givenNumbers_whenFindCombination_thenCorrect() {
    assertAll(
        () -> test(new int[]{1, 2, 3}, 4, 7)
    );
  }

  private void test(int[] given, int target, int expected) {
    // when
    Solution combinationSum4 = new Solution();
    int actual = combinationSum4.combinationSum4(given, target);

    // then
    assertEquals(expected, actual);
  }
}
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