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