LeetCode 77. Combinations
原题链接在这里:https://leetcode.com/problems/combinations/
题目:
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.
For example,
If n = 4 and k = 2, a solution is:
[ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ]
题解:
backtracking, dfs迭代的stop condition是item.size() == k, 此时把item的copy加到res中.
Time Complexity: exponential.
Space:O(k). stack space.
AC Java:
1 class Solution { 2 public List<List<Integer>> combine(int n, int k) { 3 List<List<Integer>> res = new ArrayList<List<Integer>>(); 4 if(n<=0 || k<=0 || k>n){ 5 return res; 6 } 7 dfs(n, k, 1, new ArrayList<Integer>(), res); 8 return res; 9 } 10 11 private void dfs(int n, int k, int start, List<Integer> item, List<List<Integer>> res){ 12 if(item.size() == k){ 13 res.add(new ArrayList<Integer>(item)); 14 return; 15 } 16 for(int i = start; i<=n; i++){ 17 item.add(i); 18 dfs(n, k, i+1, item, res); 19 item.remove(item.size()-1); 20 } 21 } 22 }
AC Python:
1 class Solution: 2 def combine(self, n: int, k: int) -> List[List[int]]: 3 res = [] 4 self.dfs(n, k, 1, [], res) 5 return res 6 7 def dfs(self, n: int, k: int, start: int, item: List[int], res: List[List[int]]) -> None: 8 if len(item) == k: 9 res.append(item[:]) 10 return 11 12 for i in range(start, n + 1): 13 item.append(i) 14 self.dfs(n, k, i + 1, item, res) 15 item.pop() 16
类似Combination Sum, Combination Sum II, Combination Sum III, Factor Combinations, Subsets, Permutations, N-Queens.