Combination Sum

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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.

Note:

  • All numbers (including target) will be positive integers.
  • The solution set must not contain duplicate combinations.

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

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[
  [7],
  [2, 2, 3]
]

Analysis

这道题看起来还是挺麻烦的，要搜索所有的解，对于解题来看，先简单的想一个例子，然后walk through，仔细理理应该可以想出来需要一步步的搜索加上回溯。相对来说还是比较复杂了

Solution

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class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        vector<int> out;
        vector<vector<int> > res;
        dfs(candidates, target, 0, out, res);
        return res;
    }

    void dfs(const vector<int> &candidates, int &target, int level, vector<int> &out, vector<vector<int> > &res) {
        if(target == 0) res.push_back(out);
        else {
            for (int i=level; i<candidates.size(); ++i) {
                if(target >= candidates[i]) {
                    out.push_back(candidates[i]);
                    target -= candidates[i];
                    dfs(candidates, target, i, out, res);
                    target += candidates[i];
                    out.pop_back();
                }
            }
        }
    }
};

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class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        res = []
        self.dfs(res, [], candidates, 0, target)
        return res

    def dfs (self, res, out, candidates, level, target):
        if target == 0:
            # 引用调用，最后所有的out都变成[]， 所以这里一定要deepcopy
            res.append(copy.deepcopy(out))
        else:
            for i in range (level, len(candidates)):
                if target >= candidates[i]:
                    out.append(candidates[i])
                    self.dfs(res, out, candidates, i, target-candidates[i])
                    out.pop()

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class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        res = []
        self.dfs(res, [], candidates, 0, target)
        return res

    def dfs (self, res, out, candidates, level, target):
        if target == 0:
            res.append(out)
        else:
            for i in range (level, len(candidates)):
                if target >= candidates[i]:
                    # out.append(candidates[i])
                    # out +[candidates[i]]实际上是生成了另一个list指向另一个reference，所以事实上一旦放进res之后就不会再touch
                    self.dfs(res, out + [candidates[i]], candidates, i, target-candidates[i])
                    # out.pop()

Call stack

简单的来画一个call stack，帮助理解。假设需要搜索的数组为[2,3,7], target 为 7。外层需要有一个循环来不断的找下一个元素，如果满中条件，则找到一个结果，如果不满足条件，则对元素遍历收索一遍，然后再退回来，重新开始一次新的遍历搜索 backtracking_call_stack

Python 知识点

先看例子

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a = [1,2,3]
b = a
a.append(4)
a += [5]
a = a + [6]
print('a', a)
print('b', b)

# output
a [1, 2, 3, 4, 5, 6]
b [1, 2, 3, 4, 5]

Expression a += [5] modifies the list in-place, means it extends the list such that “list1” and “list2” still have the reference to the same list.

Expression a = a + [6] creates a new list and changes “list1” reference to that new list and “list2” still refer to the old list. so here a = a + [6] makes a completely a different reference from b while previously, a and b have the reference to the same list.