Leetcode[39/40/216] combinationSum 三部曲

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# coding:utf-8
class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        def dfs(target,temp, start):
            if target == 0:
                res.append(temp)
                return
            if target > 0:
                for i in range(start, len(candidates)):
                    c = candidates[i]
                    dfs(target-c, temp+[c], i)
        res=[]
        dfs(target, [], 0)
        return res


class Solution(object):
    def combinationSum2(self, candidates, target):
        res = []
        candidates.sort()
        
        def backtrack(remain, temp, start):
            if remain==0:
                res.append(temp)
            elif remain > 0:
                for i in range(start, len(candidates)):
                    if candidates[i] > remain:
                        break
                    if i > start and candidates[i] == candidates[i-1]:
                        continue
                    n = candidates[i]
                    backtrack(remain-n, temp+[n], i+1)
        backtrack(target, [], 0)
        return res



class Solution(object):
    def combinationSum3(self, k, n):
        """
        :type k: int
        :type n: int
        :rtype: List[List[int]]
        """        
        res = []  
        # k个数的和是n,就意味着满足条件的子序列,
        # 在k = 0 and n = 0,并且是
        def dfs(k, n, tmp, start):
            if n == 0 and k == 0:
                res.append(tmp)
                return
            
            for i in range(start, 10):
                dfs(k - 1, n - i, tmp+[i], i + 1)
                    
        dfs(k, n, [], 1)       
        return res

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转载自blog.csdn.net/u010870545/article/details/88761843