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?

Analyse: For every number i from 1 ... target, we check the numbers num in nums, if i >= num, we seperate i into num and i - num, and compute how many ways i - num are consisted. That is to say, dp[i] += dp[i - num].

Runtime: 4ms. 

 1 class Solution {
 2 public:
 3     int combinationSum4(vector<int>& nums, int target) {
 4         vector<int> dp(target + 1, 0);
 5         
 6         dp[0] = 1;
 7         for(int i = 1; i <= target; i++) {
 8             for(auto num : nums) {
 9                 if(i >= num)
10                     dp[i] += dp[i - num];
11             }
12         }
13         return dp[target];
14     }
15 };

Combination Sum IV

原文：http://www.cnblogs.com/amazingzoe/p/5745334.html