回溯经典例题
https://zhuanlan.zhihu.com/p/112926891
39. 组合总和
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
def backtrack(lst,target,res,path,begin,size):
#step1:不满足条件直接返回
if target<0: return
#step2:满足条件返回结果(执行一些操作)
if target==0:
res.append(path)
return
#step3:循环遍历所有的可能性
for index in range(begin,size):
backtrack(lst,target-lst[index],res,path+[lst[index]],index,size)
size=len(candidates)
if not size: return []
res,path=[],[]
backtrack(candidates,target,res,path,0,size)
return res
140. 单词拆分 II
def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
import functools
if not wordDict:return []
wordDict = set(wordDict) #set 内部是dict实现 in操作时间复杂度O(1)
max_len = max(map(len, wordDict)) # 高级写法
@functools.lru_cache(None)
def backtrack(s):
res = []
if not s:
res.append("")
return res
for i in range(len(s)):
if i < max_len and s[:i+1] in wordDict:
for t in backtrack(s[i+1:]):
if not t:
res.append(s[:i+1])
else:
res.append(s[:i+1] + " " + t) #最后回溯的放在后面
return res
return backtrack(s)
14. 自由之路
def findRotateSteps(self, ring: str, key: str) -> int:
@lru_cache(None)
def dfs(path,s):
# 如果递归深度和key长度相同，递归结束
if path == len(key):
return 0
left = s.find(key[path])
left_str = s[left:]+s[:left]
right = s.rfind(key[path])
right_str = s[right:]+s[:right]
left_res = dfs(path+1,left_str)
right_res = dfs(path+1,right_str)
res = min(left+left_res,len(s)-right+right_res)+1
return res
return dfs(0,ring)
4. 17. 电话号码的字母组合
def letterCombinations(self, digits: str) -> List[str]:
# 回溯 时间复杂度 3**a * 4**b a,b表示字符串中对应字典值长度为3和4的数字个数
if not digits: return []
a = '23456789'
b = ['abc','def','ghi','jkl','nmo','pqrs','tuv','wxyz']
dic = dict(zip(a,b)) # 创建字典
res,state = [],[] # 结果集/中间状态集
def backtrack(idx):
if idx == len(digits):
res.append("".join(state))
else:
num = digits[idx]
for c in dic[num]:
state.append(c) # 更新状态
backtrack(idx+1) # 每次加入新字母后递归下一个位置
state.pop() # 清空状态
backtrack(0)
return res
