leetcode DFS

Summary

• DFS problems have two kinds: One to get the number of all solutions. The other is to get all the solutions itself.
  
  • To get the total number of solutions, usually you can use dynamic programming.
  • To get all the solution, there is a fixed solution style. And you need to determine where it doesn’t give out.

# main method res = []self.helper(....)return res# auxiliary methoddef helper(...):    if proper:        res.append()        return    for x in xrange(start,end):        do something        self.helper(...)        undo something

leetcode 46 Permutations

Question

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

Solution

recursion swap

Swap the current value and each of the values followed, then solve the rest permutation.

class Solution(object):    def permute(self, nums):        """        :type nums: List[int]        :rtype: List[List[int]]        """        res = []        self.helper(nums,0,res)        return res    def helper(self,nums,start,res):        if start == len(nums):            res.append(list(nums))            return        for i in xrange(start,len(nums)):            if not self.contain_duplication(nums,start,i):                nums[start],nums[i] = nums[i],nums[start]                self.helper(nums,start+1,res)                nums[start],nums[i] = nums[i],nums[start]    def contain_duplication(self,nums,start,end):        for i in xrange(start,end):            if nums[i] == nums[end]:                return True

• increase insert
  Every turn add a value in every possible position, and next turn and the new value on the base of the last turn. Use set to delete the duplication.

class Solution(object):    def permute(self, nums):        """        :type nums: List[int]        :rtype: List[List[int]]        """        res = []        res.append([])        for i in xrange(len(nums)):            s = set()            for lst in res:                for j in xrange(len(lst)+1):                    lst.insert(j,nums[i])                    s.add(tuple(lst))                    del lst[j]            res = [list(t) for t in s]        return res

leetcode 77 Combinations

Question

Given two integers n and k, return all possible combinations of k numbers out of 1 … n.

For example,
If n = 4 and k = 2, a solution is:

[
[2,4],
[3,4],
[2,3],
[1,2],
[1,3],
[1,4],
]

Solution

Recursion, when find a solution, back to the next value.

class Solution(object):    def combine(self, n, k):        """        :type n: int        :type k: int        :rtype: List[List[int]]        """        res = []        self.helper(n,k,[],res,1)        return res    def helper(self, n, k, lst, res, start):        if len(lst) == k:            res.append(list(lst))            return        for i in xrange(start,n+1):            lst.append(i)            self.helper(n,k,lst,res,i+1)            lst.pop()

leetcode 39 Combination Sum

Question

Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Solution

Different from above, this time it is allowed to reuse the elements, So you have to start at i, not i+1.

class Solution(object):    def combinationSum(self, candidates, target):        """        :type candidates: List[int]        :type target: int        :rtype: List[List[int]]        """        res = []        candidates.sort()        self.helper(candidates,0,target,[],res)        return res    def helper(self,candidates,start,target,lst,res):        if target == 0:            res.append(list(lst))            return        elif target < 0:            return        for i in xrange(start,len(candidates)):            if i == start or candidates[i] != candidates[i-1]:                 lst.append(candidates[i])                self.helper(candidates,i,target-candidates[i],lst,res)                lst.pop()

leetcode 40 Combination Sum II

Question

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.

Solution

The only difference from above is that every time it begins at the next point.

class Solution(object):    def combinationSum2(self, candidates, target):        """        :type candidates: List[int]        :type target: int        :rtype: List[List[int]]        """        res = []        candidates.sort()        self.helper(candidates,target,0,[],res)        return res    def helper(self,candidates,target,start,lst,res):        if target <= 0:            if target == 0:                res.append(list(lst))            return        for i in xrange(start,len(candidates)):            if i == start or candidates[i] != candidates[i-1]:                lst.append(candidates[i])                self.helper(candidates,target-candidates[i],i+1,lst,res)                lst.pop()

leetcode 216 Combination Sum III

Question

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Ensure that numbers within the set are sorted in ascending order.

Solution

Same to above.

class Solution(object):    def combinationSum3(self, k, n):        """        :type k: int        :type n: int        :rtype: List[List[int]]        """        res = []        self.helper(k,n,1,[],res)        return res    def helper(self,k,n,start,lst,res):        if k == 0 or n <= 0:            if k == 0 and n == 0:                res.append(list(lst))            return        for i in xrange(start,10):            lst.append(i)            self.helper(k-1,n-i,i+1,lst,res)            lst.pop()