diff --git a/problems/0077.组合.md b/problems/0077.组合.md index 9c6d481d..c4da4471 100644 --- a/problems/0077.组合.md +++ b/problems/0077.组合.md @@ -377,68 +377,42 @@ class Solution { ``` ### Python - +未剪枝优化 ```python -class Solution(object): - def combine(self, n, k): - """ - :type n: int - :type k: int - :rtype: List[List[int]] - """ - result = [] - path = [] - def backtracking(n, k, startidx): - if len(path) == k: - result.append(path[:]) - return - - # 剪枝, 最后k - len(path)个节点直接构造结果,无需递归 - last_startidx = n - (k - len(path)) + 1 - - for x in range(startidx, last_startidx + 1): - path.append(x) - backtracking(n, k, x + 1) # 递归 - path.pop() # 回溯 - - backtracking(n, k, 1) +class Solution: + def combine(self, n: int, k: int) -> List[List[int]]: + result = [] # 存放结果集 + self.backtracking(n, k, 1, [], result) return result + def backtracking(self, n, k, startIndex, path, result): + if len(path) == k: + result.append(path[:]) + return + for i in range(startIndex, n + 1): # 需要优化的地方 + path.append(i) # 处理节点 + self.backtracking(n, k, i + 1, path, result) + path.pop() # 回溯,撤销处理的节点 + ``` + +剪枝优化: + ```python class Solution: def combine(self, n: int, k: int) -> List[List[int]]: - res = [] - path = [] - def backtrack(n, k, StartIndex): - if len(path) == k: - res.append(path[:]) - return - for i in range(StartIndex, n + 1): - path.append(i) - backtrack(n, k, i+1) - path.pop() - backtrack(n, k, 1) - return res -``` + result = [] # 存放结果集 + self.backtracking(n, k, 1, [], result) + return result + def backtracking(self, n, k, startIndex, path, result): + if len(path) == k: + result.append(path[:]) + return + for i in range(startIndex, n - (k - len(path)) + 2): # 优化的地方 + path.append(i) # 处理节点 + self.backtracking(n, k, i + 1, path, result) + path.pop() # 回溯,撤销处理的节点 -剪枝: - -```python -class Solution: - def combine(self, n: int, k: int) -> List[List[int]]: - res=[] #存放符合条件结果的集合 - path=[] #用来存放符合条件结果 - def backtrack(n,k,startIndex): - if len(path) == k: - res.append(path[:]) - return - for i in range(startIndex,n-(k-len(path))+2): #优化的地方 - path.append(i) #处理节点 - backtrack(n,k,i+1) #递归 - path.pop() #回溯,撤销处理的节点 - backtrack(n,k,1) - return res ``` ### Go