[Lintcode]152. Combinations/[Leetcode]77. Combinations

Posted 2021-02-09 siriusli

152. Combinations/77. Combinations

• 本题难度: Medium
• Topic: Search & Recursion

Description

Given two integers n and k, return all possible combinations of k numbers out of 1 ... n.

Example
Given n = 4 and k = 2, a solution is:

[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4]]
Notice
You don‘t need to care the order of combinations, but you should make sure the numbers in a combination are sorted.

我的代码

class Solution:
    """
    @param n: Given the range of numbers
    @param k: Given the numbers of combinations
    @return: All the combinations of k numbers out of 1..n
    """
    def combine(self, n, k):
        # write your code here
        res = []
        s1 = [[]]
        s2 = []
        for i in range(1,n+1):
            while(s1):
                tmp1 = s1.pop()
                tmp2 = tmp1+[i]
                if len(tmp2) == k:
                    res.append(tmp2)
                else:
                    s2.append(tmp2) 
                if k-len(tmp1)<=n-i:
                    s2.append(tmp1)
            s1 = s2
            s2 = []
        return res

别人的代码

https://leetcode.com/problems/combinations/discuss/27024/1-liner-3-liner-4-liner

1. Library

from itertools import combinations

class Solution:
    def combine(self, n, k):
        return list(combinations(range(1, n+1), k))

2.Recursive - AC in 76 ms

有点像动态规划

class Solution:
    def combine(self, n, k):
        if k == 0:
            return [[]]
        return [pre + [i] for i in range(k, n+1) for pre in self.combine(i-1, k-1)]

3.Iterative - AC in 76 ms

class Solution:
    def combine(self, n, k):
        combs = [[]]
        for _ in range(k):
            combs = [[i] + c for c in combs for i in range(1, c[0] if c else n+1)]
        return combs

4. Reduce - AC in 76 ms

```python

class Solution:
def combine(self, n, k):
return reduce(lambda C, _: [[i]+c for c in C for i in range(1, c[0] if c else n+1)],
range(k), [[]])```

思路
我当时的想法是，对每个元素，有可能存在或不存在于某个结果中。其次，如果最后余下对数均加入一个结果中仍不足k，就提前舍弃。