版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/programmingfool5/article/details/82784830
DFS 一般的套路模板:
def dfs(当前状态,成员变量,全局变量)
if(当前状态为边界状态):
记录或输出
return
for i in range(横向遍历解答树所有子节点):
#扩展出一个子状态。
if(不满足条件):
continue or break
# 恢复全局变量 #回溯部分
if(子状态满足约束条件):
dfs(子状态) #修改了全局变量
题目: Permutations
Input:
[1,2,3]
Output:
[ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ]
def Permutations (nums):
ans = []
dfs(nums,[],ans)
return ans
def dfs(nums, tmp, ans):
if not nums:
#if tmp not in ans:当数组中有相同的元素时
ans.append(tmp)
return
for i in range(len(nums)):
dfs(nums[:i]+nums[i+1:], tmp+[nums[i]], ans)⭐️
题目:
Given two integers n and k, return all possible combinations of k numbers out of 1 … n. 返回k个数的组合.
Example:
Input: n = 4, k = 2
Output:
[[2,4], [3,4],[2,3], [1,2], [1,3], [1,4]]
class Solution(object):
def combine(self, n, k):
ans=[]
self.dfs(n,1,k,[],ans)
return ans
def dfs(self,n,start,k,ret,ans):
if k <= 0 :
ans.append(ret)
else:
for i in range(start,n+1):
self.dfs(n,i+1,k-1,ret+[i],ans)
题目:
子集:
Input: nums = [1,2,3]
Output:
[[3], [1], [2], [1,2,3], [1,3],[2,3], [1,2],[] ]
#1
def subset_recursive(nums):
ans = []
dfs(nums, 0, [], ans)
return ans
def dfs1(nums, start, ret, ans):
ans.append(ret)
for i in range(start, len(nums)):
dfs(nums, i + 1, ret + [nums[i]], ans)
#2 位图
class Solution(object):
def subset1(self, nums):
nums_ = pow(2,len(nums))
ans= [ []for _ in range(nums_)]
for i in range(len(nums)):
for j in range(nums_):
if (j>>i) & 1:
ans[j].append(nums[i])
return ans
带有重复元素的子集:
def subsetsWithDup(self, nums):
res = []
nums.sort() #!!! 先排序
self.dfs(nums, 0, [], res)
return res
def dfs(self, nums, index, path, res):
res.append(path)
for i in range(index, len(nums)):
if i > index and nums[i] == nums[i - 1]:⭐️
continue
self.dfs(nums, i + 1, path + [nums[i]], res)
题目:组合数的和
Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
[1,2,2],
[5]
]
class Solution:
def combinationSum2(self, nums, target):
ans = []
nums.sort()
print(nums[:])
self.dfs(nums, target, 0, [], ans)
return ret
def dfs(self, nums, target, start, path, ans):
if target == 0 :
ans.append(path)
return
for i in range(start, len(nums)):
if nums[i] > target:
break
if i > start and nums[i] == nums[i-1]:
continue
self.dfs( nums, target - nums[i], i + 1, path + [nums[i]], ans )
题目: N皇后问题
class Solution(object):
def solveNQueens(self, n):
res = []
columnforRow = [n]*n
self.dfs(n,0,columnforRow,res)
return res
def dfs(self, n, row, columnforRow,ret):
#back
if row == n:
item = []
for i in range(n):
rowstr = ""
for j in range(n):
if columnforRow[i] == j:
rowstr += 'Q'
else:
rowstr += '.'
item.append(rowstr)
ret.append(item)
return
for i in range(n):
columnforRow[row] = i ⭐️
if self.check(row,columnforRow) :
self.dfs(n,row + 1,columnforRow,ret) 子状态
#track
def check (self, row, columnforRow):
for i in range(row):
if(columnforRow[row] == columnforRow[i] or abs(columnforRow[row]-columnforRow[i]) == row - i):
return False
return True
题目:
地上有一个m行和n列的方格。一个机器人从坐标0,0的格子开始移动，每一次只能向左，右，上，下四个方向移动一格，但是不能进入行坐标和列坐标的数位之和大于k的格子。 例如，当k为18时，机器人能够进入方格（35,37），因为3+5+3+7 = 18。但是，它不能进入方格（35,38），因为3+5+3+8 = 19。请问该机器人能够达到多少个格子？
思路: 与八皇后的思路基本一致, 根据模板就可以
def moveCounts(threshold, rows, cols):
if threshold < 0 or rows<=0 or cols <= 0:
return 0
visited = [[ False for _ in range(cols)] for _ in range(rows)]
counts = self.dfs(threshold, rows, cols, 0, 0, visited)
return counts
def dfs(threshold, rows, cols, i, j, visited):
counts = 0
if self.check(threshold, rows, cols, i, j, visited):⭐️ 判定条件
visited[i][j] = True ⭐️子状态
counts = 1 + self.dfs(threshold, rows, cols, i+1, j, visited)\
+self.dfs(threshold, rows, cols, i-1, j, visited)\
+self.dfs(threshold, rows, cols, i, j+1, visited)\
+self.dfs(threshold, rows, cols, i, j-1, visited)
return counts
def check(threshold, rows, cols, i, j, visited):
if 0<= i < rows and 0<= j < cols and not visited[i][j] and self.getsum(i)+self.getsum(j)<= threshold:
return True
return False
def getsum(number):
sum_ = 0
while number > 1:
sum_ += number % 10
number //= 10
return sum _