思路

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        paths = []
        path = []
        def backtrack(n, k, start):
            if len(path) == k:
                paths.append(path[:])

            for i in range(start, n+1):
                path.append(i)
                backtrack(n, k, i+1)
                path.pop()

        backtrack(n, k, 1)
        return paths

优化一下
k-len(path）用来检测还需要多少元素，在1-n中至多要从该起始位 n - (k - len(path) + 1，开始遍历，+1代表左闭集合
距离n=k=4，则第一层为例，i最多取到1，对应结果[1,2,3,4]

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        paths = []
        path = []
        def backtrack(n, k, start):
            if len(path) == k:
                paths.append(path[:])
                return
            for i in range(start, n-(k-len(path)) + 2):
                path.append(i)
                backtrack(n, k, i+1)
                path.pop()

        backtrack(n, k, 1)
        return paths