第十周 动态规划 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.

**

思路

**
题目给定一个正整数数组和一个目标值，希望找到能由数组中的元素可重复地相加得到目标值的方案个数。
使用动态规划的思想，设s【i】表示能由数组中元素可重复相加得到i的方案个数，nums【j】表示数组中第j个元素。当i=0,1…n - 1的s【i】都求得时，则s【n】=∑ s【n - nums[j]】（nums[j] <= n）。

**源码**
--

class Solution {  
public:  
    int combinationSum4(vector<int>& nums, int target) {  
        if(nums.size() == 0)  
            return 0;  
        int s[100000],i,j,count;  
        s[0] = 1;  
        for(i = 1;i <= target;i ++){  
            count = 0;  
            for(j = 0;j < nums.size();j ++){  
                if(nums[j] <= i){  
                    count += s[i - nums[j]];  
                }  
            }  
            s[i] = count;  
        }  
        return s[target];  
    }  
};