Leetcode: 216. Combination Sum III(Week14, Medium)

Leetcode 216
Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

[[1,2,6], [1,3,5], [2,3,4]]

• 题意：定义k为组合的长度，n为目标值，现要求从数字序列1-9中，找到所有的组合(组合不重复)，且数字序列中每个数字只能使用一次，最终每个组合的和会等于目标值
• 思路：思路与Combination SumII类似，只不过是在递归函数中要多加点限制，即获得正确组合的条件是：n==0 && k == 0
• 代码如下：

class Solution {public:    vector<vector<int>> combinationSum3(int k, int n) {        vector<vector<int>> result;        vector<int> temp_result;        for (int i = 1; i <= 9; i++) {            temp_result.push_back(i);            solve_Combination_Sum(result, temp_result, k-1, n-i, i);            temp_result.pop_back();        }        return result;    }    void solve_Combination_Sum(vector<vector<int>>& result, vector<int>& out, int k, int n, int num) {        if (n < 0) return;        if (n == 0 && k == 0) result.push_back(out);        if (k == 0) return;        for (int i = num+1; i <= 9; i++) {            out.push_back(i);            solve_Combination_Sum(result, out, k-1, n-i, i);            out.pop_back();        }    }};

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