LintCode-152: Combinations (DFS 回溯)-优快云博客

本文介绍了一个经典的算法问题——从1到n中选择k个数的所有组合，并提供了详细的递归回溯解决方案。通过示例展示了输入n=4, k=2和n=4, k=1时的输出结果。

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Combinations

Given two integers n and k. Return all possible combinations of k numbers out of 1, 2, … , n.

Example
Example 1:

Input: n = 4, k = 2
Output: [[1,2],[1,3],[1,4],[2,3],[2,4],[3,4]]
Example 2:

Input: n = 4, k = 1
Output: [[1],[2],[3],[4]]
Notice
You can return all combinations in any order, but numbers in a combination should be in ascending order.

解法1：DFS，回溯。

class Solution {
public:
    /**
     * @param n: Given the range of numbers
     * @param k: Given the numbers of combinations
     * @return: All the combinations of k numbers out of 1..n
     */
    vector<vector<int>> combine(int n, int k) {
        vector<vector<int>> results;
        vector<int> sol;
        
        if ((n < k) || (k == 0)) {
            results.push_back(vector<int>());
            return results;
        }
        
        helper(n, k, 1, sol, results);
        return results;
    }

    void helper(int n, int k, int index, vector<int> &sol, vector<vector<int>> &results) {
        if (sol.size() == k) {
            results.push_back(sol);
            return;
        }

        if (index == n + 1) return;
        
        for (int i = index; i <= n; ++i) {
            sol.push_back(i);
            helper(n, k, i + 1, sol, results);
            sol.pop_back();
        }
    }    
};