Problem:
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:
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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. |
题目大意：给你一些数和一个目标值，问使用这些数(每个数可以被使用任意次数)加起来等于目标值的组合（其实是不重复的排列）有多少种。
Idea:
DP
Solution:
C++ / Recursion + Memorization
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// Author: Huahua // Runtime: 0 ms class Solution { public: int combinationSum4(vector<int>& nums, int target) { m_ = vector<int>(target + 1, -1); m_[0] = 1; return dp(nums, target); } private: int dp(const vector<int>& nums, int target) { if (target < 0) return 0; if (m_[target] != -1) return m_[target]; int ans = 0; for (const int num : nums) ans += dp(nums, target - num); return m_[target] = ans; } vector<int> m_; }; |
C++ / DP
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// Author: Huahua // Runtime: 3 ms class Solution { public: int combinationSum4(vector<int>& nums, int target) { vector<int> dp(target + 1, 0); // dp[i] # of combinations sum up to i dp[0] = 1; for (int i = 1; i <= target; ++i) for (const int num : nums) if (i - num >= 0) dp[i] += dp[i - num]; return dp[target]; } }; |
Related Problems:
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