文章目录
1.回溯算法2.回溯算法模板3.回溯实例(77、216、17、39、40、131、93、78、90、491、46、47)4.总结1.回溯算法
回溯算法的本质就是穷举,最多再加上剪枝,剪掉一部分不必要的。
关于排列组合的区别,组合无序,排列有序
回溯算法解决问题都可以抽象为树形结构(N叉树),树的宽度代表集合的大小,树的深度代表递归的深度,树的高度是有限的,也就是递归是有终止条件的。
2.回溯算法模板
void backtracking(参数) :if (终止条件) :存放结果returnfor (选择:本层集合中元素(树中节点孩子的数量就是集合的大小)):处理节点backtracking(路径,选择列表) # 递归回溯,撤销处理结果
3.回溯实例(77、216、17、39、40、131、93、78、90、491、46、47)
77. 组合 - 力扣(LeetCode) (leetcode-)
剪枝需要在for里面做处理
class Solution:def combine(self, n: int, k: int) -> List[List[int]]:def backtracking(i):if not len(path)-k: return res.append(path[:])for j in range(i,n-(k-len(path)-1)):#if j>n-(k-len(path)):break #剪枝path.append(j+1)backtracking(j+1)path.pop()res=[] #存放符合条件结果的集合path=[] #用来存放符合条件结果backtracking(0)return res
216. 组合总和 III - 力扣(LeetCode) (leetcode-)
class Solution:def combinationSum3(self, k: int, n: int) -> List[List[int]]:def backtracking(sumer,i):if not sumer and not len(path)-k:return res.append(path[:])for j in range(i,9):if sumer-j-1<0:break #剪枝path.append(j+1)backtracking(sumer-j-1,j+1)path.pop()path =[]res = []backtracking(n,0)return res
17. 电话号码的字母组合 - 力扣(LeetCode) (leetcode-)
class Solution:def letterCombinations(self, digits: str) -> List[str]:def backtracking(digits):if not digits:return res.append(''.join(path)) #如果为空开始收集数据index = int(digits[0])-2for i in nums[index]:path.append(i)backtracking(digits[1:])path.pop()nums=['abc','def','ghi','jkl','mno','pqrs','tuv','wxyz']path =[]res = []backtracking(digits)return res if digits else []
39. 组合总和 - 力扣(LeetCode) (leetcode-)
class Solution:def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:def backtacking(target,index):if not target: return res.append(path[:])for i in range(index,len(candidates)):if target-candidates[i]<0:return #剪枝path.append(candidates[i])backtacking(target-candidates[i],i)path.pop()res, path = [], []candidates.sort()backtacking(target,0)return res
40. 组合总和 II - 力扣(LeetCode) (leetcode-)
class Solution:def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:def backtacking(target,index):if not target: return res.append(path[:])for i in range(index,len(candidates)):if target-candidates[i]<0: break #剪枝if i>index and not candidates[i]-candidates[i-1]:continue #剪掉重复项path.append(candidates[i])backtacking(target-candidates[i],i+1)path.pop()res, path = [], []candidates.sort() #排序方便后面的回溯backtacking(target,0)return res
131. 分割回文串 - 力扣(LeetCode) (leetcode-)
class Solution:def partition(self, s: str) -> List[List[str]]:def backtracking(index):if not index-len(s): return res.append(path[:])for i in range(index,len(s)):selectS= s[index:i+1]if selectS!=selectS[::-1]:continuepath.append(selectS)backtracking(i+1)path.pop()res, path = [], []backtracking(0)return res
93. 复原 IP 地址 - 力扣(LeetCode) (leetcode-)
class Solution:def restoreIpAddresses(self, s: str) -> List[str]:def backtracking(index,count):if not index- len(s) and not count:return res.append('.'.join(path[:]))for i in range(index,len(s)):temp = s[index:i+1]if count<0: break #大于4个的 跳出if int(temp)>255 or len(temp)>1 and not int(temp[0]):break #判断是否为有效ippath.append(s[index:i+1])backtracking(i+1,count-1)path.pop()res, path = [], []backtracking(0,4)return res
78. 子集 - 力扣(LeetCode) (leetcode-)
不同于前面的是每次结果,它都会收集
class Solution:def subsets(self, nums: List[int]) -> List[List[int]]:def backtracking(nums):res.append(path[:])if not len(nums): return for i in range(len(nums)):path.append(nums[i])backtracking(nums[i+1:])path.pop()res, path = [], []backtracking(nums)return res
90. 子集 II - 力扣(LeetCode) (leetcode-)
class Solution:def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:def backtracking(nums):res.append(path[:])if not nums:returnfor i in range(len(nums)):if i>0 and not nums[i]-nums[i-1]:continue#剪枝path.append(nums[i])backtracking(nums[i+1:])path.pop()res, path = [], []nums.sort()backtracking(nums)return res
491. 递增子序列 - 力扣(LeetCode) (leetcode-)
class Solution:def findSubsequences(self, nums: List[int]) -> List[List[int]]:def backtracking(nums):if len(path)>1:res.append(path[:]) #只取内元素个数大于等于2if not nums:return 0# 深度遍历中每一层都会有一个全新的usage_list用于记录本层元素是否重复使用usage_list = set()for i in range(len(nums)):#递增的保证if path and nums[i]<path[-1]:continue #判断是为递增if nums[i] in usage_list:continue #去除重复项usage_list.add(nums[i])path.append(nums[i])backtracking(nums[i+1:])path.pop()res, path = [], []backtracking(nums)return res
46. 全排列 - 力扣(LeetCode) (leetcode-)
class Solution:def permute(self, nums: List[int]) -> List[List[int]]:def backstacking(nums):if not nums:return res.append(path[:]) for i in range(len(nums)):path.append(nums[i])backstacking(nums[:i]+nums[i+1:])path.pop()res, path = [], []backstacking(nums)return res
47. 全排列 II - 力扣(LeetCode) (leetcode-)
class Solution:def permuteUnique(self, nums: List[int]) -> List[List[int]]:def backtracking(nums):if not nums:return res.append(path[:])dedup= set() #去掉某一层一样的for i in range(len(nums)):if nums[i] in dedup:continuededup.add(nums[i])path.append(nums[i])backtracking(nums[:i]+nums[i+1:])path.pop()res, path= [], []backtracking(nums)return res
4.总结
还留下三道困难题,明天再看了,如果明天看完比较早,那就往下看贪心算法,希望一切顺利!
如果觉得《力扣刷题-python-回溯算法-1(回溯算法模板 题型)》对你有帮助,请点赞、收藏,并留下你的观点哦!