backtracking

问题 def subarraySum(self, nums: List[int], k: int) -> int: count = 0 target = k self.cal(nums,target,count,k) return count def cal(nums,target, count,k): if target == 0: count = count+1 target = k return count,target if target<0: return for i in range(len(nums)): self.cal(nums[:i]+nums[i+1:],target-nums[i],count,k) ''' here when the target < 0 i want to break the loop and for example if there is array 1,2,3 and my target is 2 i will go to the loop first add 1 next again add 2 which is not

问题 def subarraySum(self, nums: List[int], k: int) -> int: count = 0 target = k self.cal(nums,target,count,k) return count def cal(nums,target, count,k): if target == 0: count = count+1 target = k return count,target if target<0: return for i in range(len(nums)): self.cal(nums[:i]+nums[i+1:],target-nums[i],count,k) ''' here when the target < 0 i want to break the loop and for example if there is array 1,2,3 and my target is 2 i will go to the loop first add 1 next again add 2 which is not

问题 I'm using the backtracking algorithm described in this youtube video. Now, I should be able to get ALL possible solutions. Am I able to do this with the backtracking algoritme and how? If not possible, which other (simple) algorithm should I use? 回答1: This question is not a great fit for this site since it does not appear to be about actual code. But I'll take a shot at it anyways. Of course you can get all possible solutions with a backtracking algorithm. Remember how a backtracking

问题 I'm using the backtracking algorithm described in this youtube video. Now, I should be able to get ALL possible solutions. Am I able to do this with the backtracking algoritme and how? If not possible, which other (simple) algorithm should I use? 回答1: This question is not a great fit for this site since it does not appear to be about actual code. But I'll take a shot at it anyways. Of course you can get all possible solutions with a backtracking algorithm. Remember how a backtracking

问题 I need to write a recusive function that recive an integer num and returns the number of solutions to the equation : x1 + x2 + x3 = num , where x1,x2,x3 are numbers between 1-10, the method should print all solutions. For example if num=3 then the method will print 1+1+1 and will return 1 . if num=5 the method will return 6 and will print: 1 + 1 + 3 1 + 2 + 2 1 + 3 + 1 2 + 1 + 2 2 + 2 + 1 3 + 1 + 1 if num<3 or num>30 the method will return 0 . The method should be recursive without using

问题 I need to write a recusive function that recive an integer num and returns the number of solutions to the equation : x1 + x2 + x3 = num , where x1,x2,x3 are numbers between 1-10, the method should print all solutions. For example if num=3 then the method will print 1+1+1 and will return 1 . if num=5 the method will return 6 and will print: 1 + 1 + 3 1 + 2 + 2 1 + 3 + 1 2 + 1 + 2 2 + 2 + 1 3 + 1 + 1 if num<3 or num>30 the method will return 0 . The method should be recursive without using

问题 I need to write a recusive function that recive an integer num and returns the number of solutions to the equation : x1 + x2 + x3 = num , where x1,x2,x3 are numbers between 1-10, the method should print all solutions. For example if num=3 then the method will print 1+1+1 and will return 1 . if num=5 the method will return 6 and will print: 1 + 1 + 3 1 + 2 + 2 1 + 3 + 1 2 + 1 + 2 2 + 2 + 1 3 + 1 + 1 if num<3 or num>30 the method will return 0 . The method should be recursive without using

问题 I'm trying to understand the semicolon functionality. I have this code: del(X,[X|Rest],Rest). del(X,[Y|Tail],[Y|Rest]) :- del(X,Tail,Rest). permutation([],[]). permutation(L,[X|P]) :- del(X,L,L1), permutation(L1,P). It's the simple predicate to show all permutations of given list. I used the built-in graphical debugger in SWI-Prolog because I wanted to understand how it works and I understand for the first case which returns the list given in argument. Here is the diagram which I made for

问题 I'm trying to understand the semicolon functionality. I have this code: del(X,[X|Rest],Rest). del(X,[Y|Tail],[Y|Rest]) :- del(X,Tail,Rest). permutation([],[]). permutation(L,[X|P]) :- del(X,L,L1), permutation(L1,P). It's the simple predicate to show all permutations of given list. I used the built-in graphical debugger in SWI-Prolog because I wanted to understand how it works and I understand for the first case which returns the list given in argument. Here is the diagram which I made for

问题 I'm trying to understand the semicolon functionality. I have this code: del(X,[X|Rest],Rest). del(X,[Y|Tail],[Y|Rest]) :- del(X,Tail,Rest). permutation([],[]). permutation(L,[X|P]) :- del(X,L,L1), permutation(L1,P). It's the simple predicate to show all permutations of given list. I used the built-in graphical debugger in SWI-Prolog because I wanted to understand how it works and I understand for the first case which returns the list given in argument. Here is the diagram which I made for