LC216.组合总和III:
class Solution:
def __init__(self):
self.res = []
self.sum = 0
self.path = []
def combinationSum3(self, k: int, n: int) -> List[List[int]]:
self.backtracking(k, n, 1)
return self.res
def backtracking(self, k: int, n: int, start_num: int):
if self.sum > n: # 剪枝
return
if len(self.path) == k: # len(path)==k时不管sum是否等于n都会返回
if self.sum == n:
self.res.append(self.path[:])
return
for i in range(start_num, 10 - (k - len(self.path)) + 1):
self.path.append(i)
self.sum += i
self.backtracking(k, n, i+1)
self.path.pop()
self.sum -= i
LC17.电话号码的字母组合
class Solution:
def __init__(self):
self.answers: List[str] = []
self.answer: str = ''
self.letter_map = {
'2': 'abc',
'3': 'def',
'4': 'ghi',
'5': 'jkl',
'6': 'mno',
'7': 'pqrs',
'8': 'tuv',
'9': 'wxyz'
}
def letterCombinations(self, digits: str) -> List[str]:
self.answers.clear()
if not digits: return []
self.backtracking(digits, 0)
return self.answers
def backtracking(self, digits: str, index: int) -> None:
# 回溯函数没有返回值
# Base Case
if index == len(digits): # 当遍历穷尽后的下一层时
self.answers.append(self.answer)
return
# 单层递归逻辑
letters: str = self.letter_map[digits[index]]
for letter in letters:
self.answer += letter # 处理
self.backtracking(digits, index + 1) # 递归至下一层
self.answer = self.answer[:-1] # 回溯