377. Combination Sum IV(unsolved)
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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?
解答:
class Solution {public: int combinationSum4(vector<int>& nums, int target) { vector<int> dp(target+1,0); dp[0]=1; sort(nums.begin(),nums.end()); for(int i=1;i<=target;i++) { for(int num:nums) { if(i<num) break; dp[i]+=dp[i-num]; } } return dp.back(); }};
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- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV**
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
- 377. Combination Sum IV
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