LeetCode 39. Combination Sum

[Woodyiiiiiii]

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]
]

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这道题的常见解法是DFS+回溯，跟做过的permutations, subset等题目一样的思路，主要是要注意递归调用时各个变量的含义就可以了，不再赘述：

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        dfs(res, tmp, candidates, target, 0, 0);
        return res;
    }
    public void dfs(List<List<Integer>> res, List<Integer> tmp,
                   int[] candidates, int target, int sum, int level) {
        if (sum > target) return;
        else if (sum == target) {
            res.add(new ArrayList<>(tmp));
            return;
        }
        for (int i = level; i < candidates.length; ++i) {
            tmp.add(candidates[i]);
            sum += candidates[i];
            dfs(res, tmp, candidates, target, sum, i);
            sum -= tmp.get(tmp.size() - 1);
            tmp.remove(tmp.size() - 1);
        }
    }
}

第二种递归写法是不用sum这个变量来计算，直接将target减去相应的值作为参数传递：

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        dfs(0, res, tmp, candidates, target);  
        return res;
    }
    
    public void dfs(int level, List<List<Integer>> res, List<Integer> tmp,
                    int[] candidates, int target){
        if (target < 0) return;
        if (target == 0) {
            res.add(new ArrayList<>(tmp));
            return;
        }
        for(int i = level ; i < candidates.length ; ++i){
            tmp.add(candidates[i]);
            dfs(i, res, tmp, candidates, target - candidates[i] );
            tmp.remove(tmp.size()-1);
        }
    }
}

此外可以将List类型的tmp变量改位Stack类，会发现运行时间会小很多。

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参考资料：

• https://leetcode.com/problems/combination-sum/
• https://leetcode.com/problems/combination-sum/discuss/600006/Java-oror-Faster-than-73-oror-3-ms-oror-Backtracking