【回溯】【leetcode题解】【M】【57】Combination Sum

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.

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.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

Subscribe to see which companies asked this question

详情见注释，两种写法，同一个思路

class Solution(object):
    count = 0

    '''
    def combinationSum(self, candidates, target):
        res = []
        candidates.sort()
        self.dfs(candidates, target, 0, [], res)
        return res

    def dfs(self, nums, target, index, path, res):
        #self.count += 1
        #print self.count
        if target < 0:
            return  # backtracking
        if target == 0:
            res.append(path)
            return
        for i in xrange(index, len(nums)):
            self.dfs(nums, target-nums[i], i, path+[nums[i]], res)

    '''

    def __init__(self):
        self.res = []
    def bt(self,c,temp,start,total,target):
        #print temp
        #self.count += 1
        #print self.count
        if total == target:
            self.res.append(temp)
            return
            #temp.sort()
            #if temp not in self.res:
            #    self.res.append(temp)
        else:
            for i in range(start,len(c)):
                if total + c[i] <= target:
                    #temp.append(c[i])#一个是，用append会非常慢，而且必须要用temp[:]指定，否则传值会出现错误
                    self.bt(c,temp+[c[i]],i,total+c[i],target)
                    #这里非常重要，这里面的start要改成i，不能每次都从零开始，否则会出现重复
                    #temp.pop(-1)#可能是，因为，这个不需要往回走，只需要往前看就行了，所以不用pop

    def combinationSum(self, candidates, target):

        c = candidates
        c.sort()
        #res = []
        #temp = []

        self.bt(c,[],0,0,target)
        self.res.sort()

        #print res
        return self.res