LeetCode: Combination Sum IV

最新推荐文章于 2025-02-03 15:37:10 发布

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最新推荐文章于 2025-02-03 15:37:10 发布

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分类专栏： LeetCode 文章标签： LeetCode dynamic programming combination sum IV

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本文探讨了给定正整数数组求和至目标值的所有可能组合数量问题，并提供了一个具体的解决方案。通过示例展示了不同序列被视为不同组合的情况，并讨论了允许负数时问题的变化。

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Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

Example:

nums = [1, 2, 3]
target = 4

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.

Therefore the output is 7.

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?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.

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class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        vector<int> selected;
        for (int i = 0; i < nums.size(); ++i) {
            if (nums[i] <= target) {
                selected.push_back(nums[i]);
            }
            else {
                break;
            }
        }
        if (selected.size() == 0) {
            return 0;
        }
        
        vector<int> dp = vector<int>((target + 1), 0);
        dp[0] = 1;
        
        for(int i = selected[0]; i <= target; ++i)
        {
            for (int j = 0; j < selected.size(); ++j)
            {
                if (i < selected[j]) {
                    break;
                }
                
                dp[i] += dp[i - selected[j]];
            }
        }
        
        return dp[target];
    }
};