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:

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(n<=0 || k<=0 || k>n){
            return res;
        } 
        dfs(n, k, 1, new ArrayList<Integer>(), res);
        return res;
    }
    
    private void dfs(int n, int k, int start, List<Integer> item, List<List<Integer>> res){
        if(item.size() == k){
            res.add(new ArrayList<Integer>(item));
            return;
        }
        for(int i = start; i<=n; i++){
            item.add(i);
            dfs(n, k, i+1, item, res);
            item.remove(item.size()-1);
        }
    }
}

AC Python:

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        res = []
        self.dfs(n, k, 1, [], res)
        return res
    
    def dfs(self, n: int, k: int, start: int, item: List[int], res: List[List[int]]) -> None:
        if len(item) == k:
            res.append(item[:])
            return
        
        for i in range(start, n + 1):
            item.append(i)
            self.dfs(n, k, i + 1, item, res)
            item.pop()
        

类似Combination Sum, Combination Sum II, Combination Sum III, Factor Combinations, Subsets, Permutations, N-Queens.