[LeetCode] 39. Combination Sum 组合之和

Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

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

Note:

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

Example 1:

Input: candidates = [2,3,6,7], target = 7,
A solution set is:
[
  [7],
  [2,2,3]
]

Example 2:

Input: candidates = [2,3,5], target = 8,
A solution set is:
[
  [2,2,2,2],
  [2,3,3],
  [3,5]
]

对于一个给出的数target和一个数集C，问存在多少种不同的方案，使得可以从数集C中选出若干个数（每个数可以选择无限次）使得这些数的和等于target。

新题中没有：

• Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).

如果有这个要求就要对数组C排序，使用过的就不能在使用了，因为会要保证大小顺序。

解法：递归 + backtracking,  注意数组的mutable.

Java：

public class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        if (candidates == null) return res;
        Arrays.sort(candidates);
        getCombinations(res, new ArrayList<>(), candidates, target, 0);
        return res;
    }
    
    private void getCombinations(List<List<Integer>> res, List<Integer> list, int[] nums, int target, int pos) {
        if (target < 0) return;
        if (target == 0) {
            res.add(new ArrayList<>(list));
            return;
        }
        for (int i = pos; i < nums.length; i++) {
            if (nums[i] > target) break;
            if (i > pos && nums[i] == nums[i - 1]) continue;
            list.add(nums[i]);
            getCombinations(res, list, nums, target - nums[i], i);
            list.remove(list.size() - 1);
        }
    }
}

Python:

class Solution:
    # @param candidates, a list of integers
    # @param target, integer
    # @return a list of lists of integers
    def combinationSum(self, candidates, target):
        result = []
        self.combinationSumRecu(sorted(candidates), result, 0, [], target)
        return result
    
    def combinationSumRecu(self, candidates, result, start, intermediate, target):
        if target == 0:
            result.append(list(intermediate))
        while start < len(candidates) and candidates[start] <= target:
            intermediate.append(candidates[start])
            self.combinationSumRecu(candidates, result, start, intermediate, target - candidates[start])
            intermediate.pop()
            start += 1　　

Python: wo

class Solution(object):
    def combinationSum(self, candidates, target):
        """
        :type candidates: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        res = []        
        self.helper(res, sorted(candidates), [], 0, 0, target)      
        return res
    
    def helper(self, res, candidates, cur, temp, pos, target):
        if temp > target:
            return
        if temp == target:
            res.append(cur)
            
        for i in range(pos, len(candidates)):
            self.helper(res, candidates, cur + [candidates[i]], temp + candidates[i], i, target)　　

C++:

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
        // 注意candidates不是有序的，要先排序处理
        sort(candidates.begin(), candidates.end());
        // 用以记录答案和当前方案
        vector<vector<int>> ans;
        vector<int> current;
        // 开始回溯搜索
        backtracking(ans, candidates, current, 0, target);
        // 返回答案
        return ans;
    }
    
    void backtracking(vector<vector<int>>& ans, vector<int>& candidates, vector<int> current, int last_use, int rest_target) {
        // 当rest_target等于0时，说明已经找到了一组合法的方案
        if (rest_target == 0) {
            // 故将其加入到答案当中
            ans.push_back(current);
        }
        // 否则就枚举下一个加入到current中的数，在枚举中注意2个条件
        // i >= last_use，保证current是非递减的
        // candidates[i] <= rest_target，保证rest_target不小于0
        for (int i = last_use; i < candidates.size() && candidates[i] <= rest_target; i++) {
            // 放入current中
            current.push_back(candidates[i]);
            // 继续搜索下一个数字
            backtracking(ans, candidates, current, i, rest_target - candidates[i]);
            // 回溯处理
            current.pop_back();
        }
    }
};

　　　

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