Combination Sum IV

原题链接： https://leetcode.com/problems/combination-sum-iv/description/

原题： 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.

样例：
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.

Solution： 此题可以采用动态规划的解法，用combinations[i]表示用给定数组nums能组成i的方式数量。这样，combinations[0]就应为1，只要我们计算出combinations[target]，那么它就是要求的解。

此问题的状态转移可表示为如下形式：
combinations[0] = 1;
combinations[i] = combinations[i - nums[0]] + combinations[i - nums[1]] + … + combinations[i - nums[n]]; 其中，i - nums[j]的大小必须大于等于0；

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

复杂度： 设数组nums的大小为n，那么此算法的时间复杂度为O(n * target），空间复杂度为O(target)。