题目：
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 = 4The 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?
方法一：回溯法 TLE
1.可以无限次使用元素
2.不同排列算作不同
3.没有重复元素
4.返回组合个数
class Solution {
public:vector<vector<int> > res;void tracebackComSum(vector<int> &nums, vector<int> &comb, int target){if(target <=0 ){if(target == 0)res.push_back(comb);return;}for(int i = 0; i < nums.size(); i++){comb.push_back(nums[i]);tracebackComSum(nums,comb,target-nums[i]);comb.pop_back();}}int combinationSum4(vector<int>& nums, int target) {vector<int> Comb;res.clear();tracebackComSum(nums,Comb,target);return res.size();}
};
方法二：动态规划
testcase [3,33,333] 10000 出现以上错误，把 int 改为unsigned int
class Solution {
public:int combinationSum4(vector<int>& nums, int target) {vector<unsigned int> dp(target + 1);dp[0] = 1;sort(nums.begin(),nums.end());for(int i = 1; i <= target; i++){for(auto num:nums){if(i < num)break;dp[i] = dp[i] + dp[i - num];}}return dp.back();}
};
