A typical backtracking problem. For any backtracking problem, you need to be think about three ascepts:
- What is a partial solution and when is it finished? --- In this problem, the partial solution is a combination (sol). It is finished once it contains k elements.
- How to find all the partial solutions? --- In the following code, I simply traverse all the possible starting elements (start).
- When to make recursive calls? --- In the following code, once an element is added to the partial solution, we call the function to generate the remaining elements recursively.
Of course, remember to recover to the previous status once a partial solution is done. In the following code, line 18 (sol.pop_back()) is for this purpose.
The following should be self-explanatory :)
1 class Solution { 2 public: 3 vector<vector<int>> combine(int n, int k) { 4 vector<vector<int> > res; 5 vector<int> sol; 6 combination(1, n, k, sol, res); 7 return res; 8 } 9 private: 10 void combination(int start, int n, int k, vector<int>& sol, vector<vector<int> >& res) { 11 if (k == 0) { 12 res.push_back(sol); 13 return; 14 } 15 for (int i = start; i <= n; i++) { 16 sol.push_back(i); 17 combination(i + 1, n, k - 1, sol, res); 18 sol.pop_back(); 19 } 20 } 21 };
本文探讨了一种典型的回溯法问题——从1到n中选取k个数的所有可能组合。通过定义部分解及其完成条件、寻找所有部分解的方法及递归调用时机,详细解析了算法实现过程。

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