Combination Sum IV 组合数

声明：原题目转载自LeetCode，解答部分为原创

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:

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?

Solution :

         思路：第二类的背包问题，相较于第一类，每次的操作不再是“针对某一个特定的数值，选择放入或者不放入”，而是“针对所有的数值，选择哪些数值放入，哪些数值不放入”。同时，根据题意，数量与放入的顺序有关，因此应该把这个问题当成排列问题来看待。

         假定f（a）为数值之和为a的排列的种类数，则状态转化方程为 f（a） = ∑（f（a - array[ i ]））,其中 i 的取值为0,1,2,...n

         代码如下：

#include<iostream>#include<vector>using namespace std;class Solution {public:    int combinationSum4(vector<int>& nums, int target) {        vector<int> count_of_sum(target + 1, 0);        count_of_sum[0] = 1;                for(int i = 1; i <= target ; i ++)        {        for(int j = 0 ; j < nums.size(); j ++)        {        if(i - nums[j] >= 0)        count_of_sum[i] += count_of_sum[i - nums[j]];}}return count_of_sum[target];    }};int main(){Solution text;vector<int> temp(3,0);for(int i = 0 ; i < 3; i ++)temp[i] = i + 1;cout << text.combinationSum4(temp, 4) << endl;return 0;}