leetcode 377. Combination Sum IV

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.

题目大意：找出数组中能组合成目标数的组合， 每个数字的使用次数不限

思路：用dp算法：从底到顶的计算，依次算出组合成1,2,3,4有多少种组合；

　　以上面的例子为例：【1，2，3】 target=4;

　　设置一个数组dp()来保存结果， dp[i]表示和为i的组合个数， 初始条件是dp[0]=1;

       遍历数组如果当前的数num[j]比target小那么,只需知道 dp[target-num[j]] 就行了。相当于在和为target-nums[j]的组合后面构成了新的组合，满足和为target， 把所有满足nums[j]<target的dp[target-nums[j]]加起来就得到了target的组合； 

       dp的代码简洁，关键在于找到关系式

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