Leetcode回溯算法模板
回溯算法Leetcode汇总：
78.子集
class Solution:
def subsets(self, nums):
if not nums:
return []
res = []
n = len(nums)
def helper(idx, temp_list):
res.append(temp_list)
for i in range(idx, n):
helper(i + 1, temp_list + [nums[i]])
helper(0, [])
return res
90.子集2
class Solution(object):
def subsetsWithDup(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return []
n = len(nums)
res = []
nums.sort()
# 思路1
def helper1(idx, n, temp_list):
if temp_list not in res:
res.append(temp_list)
for i in range(idx, n):
helper1(i + 1, n, temp_list + [nums[i]])
# 思路2
def helper2(idx, n, temp_list):
res.append(temp_list)
for i in range(idx, n):
if i > idx and nums[i] == nums[i - 1]:
continue
helper2(i + 1, n, temp_list + [nums[i]])
helper2(0, n, [])
return res
46.全排列
class Solution(object):
def permute(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return
res = []
n = len(nums)
visited = [0] * n
def helper1(temp_list,length):
if length == n:
res.append(temp_list)
for i in range(n):
if visited[i] :
continue
visited[i] = 1
helper1(temp_list+[nums[i]],length+1)
visited[i] = 0
def helper2(nums,temp_list,length):
if length == n:
res.append(temp_list)
for i in range(len(nums)):
helper2(nums[:i]+nums[i+1:],temp_list+[nums[i]],length+1)
helper1([],0)
return res
47.全排列2
class Solution(object):
def permuteUnique(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
if not nums:
return []
nums.sort()
n = len(nums)
visited = [0] * n
res = []
def helper1(temp_list, length):
# if length == n and temp_list not in res:
# res.append(temp_list)
if length == n:
res.append(temp_list)
for i in range(n):
if visited[i] or (i > 0 and nums[i] == nums[i - 1] and not visited[i - 1]):
continue
visited[i] = 1
helper1(temp_list + [nums[i]], length + 1)
visited[i] = 0
def helper2(nums, temp_list, length):
if length == n and temp_list not in res:
res.append(temp_list)
for i in range(len(nums)):
helper2(nums[:i] + nums[i + 1:], temp_list + [nums[i]], length + 1)
helper1([],0)
# helper2(nums, [], 0)
return res
39.组合总和
class Solution(object):
def combinationSum(self, candidates, target):
"""
:type candidates: List[int]
:type target: int
:rtype: List[List[int]]
"""
if not candidates:
return []
if min(candidates) > target:
return []
candidates.sort()
res = []
def helper(candidates, target, temp_list):
if target == 0:
res.append(temp_list)
if target < 0:
return
for i in range(len(candidates)):
if candidates[i] > target:
break
helper(candidates[i:], target - candidates[i], temp_list + [candidates[i]])
helper(candidates,target,[])
return res
40.组合总和2
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
if not candidates:
return []
candidates.sort()
n = len(candidates)
res = []
def backtrack(i, tmp_sum, tmp_list):
if tmp_sum == target:
res.append(tmp_list)
return
for j in range(i, n):
if tmp_sum + candidates[j] > target : break
if j > i and candidates[j] == candidates[j-1]:continue
backtrack(j + 1, tmp_sum + candidates[j], tmp_list + [candidates[j]])
backtrack(0, 0, [])
return res
总结：回溯算法很标准的模板
class Solution:
def combinationSum2(self, candidates: List[int], target: int) -> List[List[int]]:
if not candidates:
return []
candidates.sort()
n = len(candidates)
res = []
def backtrack(i, tmp_sum, tmp_list):
if tmp_sum == target:
res.append(tmp_list)
return
for j in range(i, n):
if tmp_sum + candidates[j] > target : break
if j > i and candidates[j] == candidates[j-1]:continue
backtrack(j + 1, tmp_sum + candidates[j], tmp_list + [candidates[j]])
backtrack(0, 0, [])
return res