0377. Combination Sum IV (M)

Combination Sum IV (M)

题目

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

The answer is guaranteed to 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?

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题意

给定一个数组，每次可以从中任取一个数，使得到的序列的和正好为指定值，求这样的序列的个数。

思路

动态规划。dp[i]表示target为i时满足的序列的数量，则遍历数组中每一个数num，有递推式(dp[i]=sum{dp[i-num]})。

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代码实现

Java

class Solution {
    public int combinationSum4(int[] nums, int target) {
        int[] dp = new int[target + 1];

        Arrays.sort(nums);
        dp[0] = 1;

        for (int i = 1; i <= target; i++) {
            for (int num : nums) {
                if (i < num) break;
                dp[i] += dp[i - num];
            }
        }
        
        return dp[target];
    }
}