回溯--组合型

 

 

 例题：

方法一：枚举下一个数选哪个

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        ans = []
        path = []
        def dfs(i: int) -> None:
            d = k - len(path)  # 还要选 d 个数
            if d == 0:  # 选好了
                ans.append(path.copy())
                return
            for j in range(i, d - 1, -1):
                path.append(j)
                dfs(j - 1)
                path.pop()  # 恢复现场
        dfs(n)
        return ans

方法二：选或不选

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        ans = []
        path = []
        def dfs(i: int) -> None:
            d = k - len(path)  # 还要选 d 个数
            if d == 0:  # 选好了
                ans.append(path.copy())
                return
            # 如果 i > d，可以不选 i
            if i > d: dfs(i - 1)
            # 选 i
            path.append(i)
            dfs(i - 1)
            path.pop()  # 恢复现场
        dfs(n)
        return ans

 

 

 

 

 

class Solution:
    def generateParenthesis(self, n: int) -> List[str]:
        m = n*2
        ans = []
        path = [''] * m
        

        def dfs(i, open):

            if i == m:
                ans.append(''.join(path))
                return 
            
            if open < n:
                path[i] = '('
                dfs(i+1, open+1)
            if i-open < open:
                path[i] = ')'
                dfs(i+1, open)
        
        dfs(0,0)
        return ans