标签：span com ble possible number bin backtrack for ++
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
Example:
Input: n = 4, k = 2 Output: [ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ]
public class Solution { public List<List<Integer>> combine(int n, int k) { List<List<Integer>> result = new ArrayList(); List<Integer> tmp = new ArrayList(); dfs(n, k, 1, 0, tmp, result); return result; } // start，开始的数, cur，已经选择的数目 private static void dfs(int n, int k, int start, int cur, List<Integer> tmp, List<List<Integer>> result) { if (cur == k) { result.add(new ArrayList(tmp)); } for (int i = start; i <= n; ++i) { tmp.add(i); dfs(n, k, i + 1, cur + 1, tmp, result); tmp.remove(tmp.size() - 1); } } }
backtracking的变种
标签：span com ble possible number bin backtrack for ++
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11538277.html
