剑指 Offer II 080. 含有 k 个元素的组合

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剑指 Offer II 080. 含有 k 个元素的组合

题目描述

给定两个整数 n 和 k，返回 1 ... n 中所有可能的 k 个数的组合。

 

示例 1:

输入: n = 4, k = 2
输出:
[
  [2,4],
  [3,4],
  [2,3],
  [1,2],
  [1,3],
  [1,4],
]

示例 2:

输入: n = 1, k = 1
输出: [[1]]

 

提示:

• 1 <= n <= 20
• 1 <= k <= n

 

注意：本题与主站 77 题相同： https://leetcode.cn/problems/combinations/

解法

方法一

Python3

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        res=[]
        path=[]
        def dfs(i):
            if len(path)==k:
                res.append(path[:])
                return
            
            for j in range(i,n+1): #层选法
                path.append(j)
                dfs(j+1)
                path.pop()
        dfs(1)
        return res

Java

class Solution {
    public List<List<Integer>> combine(int n, int k) {
        List<List<Integer>> res = new ArrayList<>();
        dfs(1, n, k, new ArrayList<>(), res);
        return res;
    }

    private void dfs(int i, int n, int k, List<Integer> t, List<List<Integer>> res) {
        if (t.size() == k) {
            res.add(new ArrayList<>(t));
            return;
        }
        for (int j = i; j <= n; ++j) {
            t.add(j);
            dfs(j + 1, n, k, t, res);
            t.remove(t.size() - 1);
        }
    }
}

C++

class Solution {
public:
    vector<vector<int>> combine(int n, int k) {
        vector<vector<int>> res;
        vector<int> t;
        dfs(1, n, k, t, res);
        return res;
    }

    void dfs(int i, int n, int k, vector<int> t, vector<vector<int>>& res) {
        if (t.size() == k) {
            res.push_back(t);
            return;
        }
        for (int j = i; j <= n; ++j) {
            t.push_back(j);
            dfs(j + 1, n, k, t, res);
            t.pop_back();
        }
    }
};

Go

func combine(n int, k int) [][]int {
	var res [][]int
	var t []int
	dfs(1, n, k, t, &res)
	return res
}

func dfs(i, n, k int, t []int, res *[][]int) {
	if len(t) == k {
		*res = append(*res, slices.Clone(t))
		return
	}
	for j := i; j <= n; j++ {
		t = append(t, j)
		dfs(j+1, n, k, t, res)
		t = t[:len(t)-1]
	}
}

Swift

class Solution {
    func combine(_ n: Int, _ k: Int) -> [[Int]] {
        var res = [[Int]]()
        dfs(1, n, k, [], &res)
        return res
    }

    private func dfs(_ start: Int, _ n: Int, _ k: Int, _ current: [Int], _ res: inout [[Int]]) {
        if current.count == k {
            res.append(current)
            return
        }

        if start > n {
            return
        }

        for i in start...n {
            var newCurrent = current
            newCurrent.append(i)
            dfs(i + 1, n, k, newCurrent, &res)
        }
    }
}