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],
]

是一种穷举遍历的思想；

自己的代码：（递归 DFS）

Runtime: 39 ms, faster than 37.50% of Java online submissions for Combinations.

Memory Usage: 43.8 MB, less than 26.50% of Java online submissions for Combinations.

import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;class Solution {public static List<List<Integer>> combine(int n, int k) {List<List<Integer>> results = new ArrayList<>();List<Integer> tempList = new ArrayList<>();  helper(results,tempList,1,n,k);return results;}public static void helper(List<List<Integer>> results ,List<Integer> tempList,int start,int n, int k) {if(k == 0) {List<Integer> list = new ArrayList<>();for(Integer i : tempList) {list.add(i);}results.add(list);return;}for(int i=start;i<=n;i++) {tempList.add(i);// not add(start)helper(results,tempList,i+1,n,k-1); // 参数三：是i+1，不是start+1tempList.remove(new Integer(i));}}public static void main(String[] args) {int n = 4;int k = 2;List<List<Integer>> list = combine(n,k);System.out.println(list);}
}

递归思想：（分解成子问题）

0）因为不确定k的大小，所以想到用递归

1）helper(start,n,k)的功能就是获得从start到n的数，组成的长度为k的所有可能结果

2）那么helper(start+1,n,k-1)的功能就是获得从start+1到n的数，组成的长度为k-1的所有可能结果

自己的代码2

import java.util.ArrayList;
import java.util.List;public class WelcomeApp {private static int count = 0;private static StringBuilder sb = new StringBuilder();private static List<String> lists = new ArrayList<>();// 获得n选k的全部组合public static void getK(int n, int k, int index,int start) {if(index > 3) {lists.add(sb.toString());count++;return;}for(int i=start;i<=n-k+1;i++) {String temp = i+" ";sb.append(temp);getK(n, k-1, index+1,i+1); // 注意第2 3 4个参数的设置；sb.delete(sb.length()-temp.length(), sb.length()); // 这句回溯一定不能少}}public static void main(String[] args) {int n=12;// 1,2,3,4,5int k=3;//取出3个数getK(n, k, 1,1);for(String str : lists) { System.out.println(str);}System.out.println("count = "+count);}}

小结：

a）应用DFS进行穷举遍历时，回溯的步骤是一定不能少的

别人的代码：

方法一：递归 + 回溯（和我的思路一模一样）

class Solution {public static List<List<Integer>> combine(int n, int k) {List<List<Integer>> combs = new ArrayList<List<Integer>>();combine(combs, new ArrayList<Integer>(), 1, n, k);return combs;}public static void combine(List<List<Integer>> combs, List<Integer> comb, int start, int n, int k) {if(k==0) {combs.add(new ArrayList<Integer>(comb)); // 直接通过构造函数，复制一个listreturn;}for(int i=start;i<=n;i++) {comb.add(i);combine(combs, comb, i+1, n, k-1);comb.remove(comb.size()-1);  // 使用的是remove(index)这个方法}}
}

1）时间复杂度：The time complexity is "n times n choose k".

"n choose k" equals (n!/((n-k)! * k!)). In full notation is O(n * (n!/((n-k)! * k!))).

We will have "n choose k" combinations and in each call we will do O(n) work to copy the array to the answer.

和方法一相同的思路的Python解法：

class Solution(object):def combine(self, n, k):""":type n: int:type k: int:rtype: List[List[int]]"""res = []def gen(i, tmp):if len(tmp) == k:res.append(list(tmp))returnif i > n:returnfor j in range(i, n+1):tmp.append(j)gen(j+1, tmp)tmp.pop()gen(1, [])return res

class Solution(object):def combine(self, n, k):""":type n: int:type k: int:rtype: List[List[int]]"""results = []result = []def DFS(results, result, start, n, k):if(k==len(result)):results.append(copy.deepcopy(result))  # python append()与深拷贝、浅拷贝应该注意，此处有一个大坑return  for i in range(start,n+1,1):result.append(i)DFS(results, result, i+1, n, k)result.pop()DFS(results, result, 1, n, k)return results

"""
[Python]_DFS_beats 97.24%The key is to stop earlier (i.e. prune the search space) if the number of numbers could be selected (n-i) is less than the number of numbers we need to select (k).当剩余的可以被选择的数的个数n-i，小于我们所需要选择的数的个数k，那么就提前终止，不用递归函数再进行下去
"""class Solution:def combine(self, n: 'int', k: 'int') -> 'List[List[int]]':res = []cur = []self.combine_helper(n, k, 0, cur, res)return resdef combine_helper(self, n, k, idx, cur, res):if k == 0:res.append(cur[:])else:for i in range(idx, n):if k > (n-i):breakval = i + 1cur.append(val)self.combine_helper(n, k-1, i+1, cur, res)cur.pop()

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自己的代码：（递归 DFS）

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