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?
Analyse: For every number i from 1 ... target, we check the numbers num in nums, if i >= num, we seperate i into num and i - num, and compute how many ways i - num are consisted. That is to say, dp[i] += dp[i - num].
Runtime: 4ms.
1 class Solution { 2 public: 3 int combinationSum4(vector<int>& nums, int target) { 4 vector<int> dp(target + 1, 0); 5 6 dp[0] = 1; 7 for(int i = 1; i <= target; i++) { 8 for(auto num : nums) { 9 if(i >= num) 10 dp[i] += dp[i - num]; 11 } 12 } 13 return dp[target]; 14 } 15 };
时间: 2024-10-16 20:16:40