【LeetCode】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] 

java code : 这种递归会超时，分析原因可能是因为频繁压栈出栈，因为每一次递归有两个递归分支，换成循环迭代版本的就不会超时了：

以下代码超时：

public class Solution {    ArrayList<ArrayList<Integer>> res;    ArrayList<Integer> newcandidates;    ArrayList<Integer> sum;    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        res = new ArrayList<ArrayList<Integer>>();if(candidates.length == 0)return res;newcandidates = new ArrayList<Integer>();Arrays.sort(candidates);for(int i = 0; i < candidates.length; i++){int num = target / candidates[i];for(int k = 0; k < num; k++)newcandidates.add(candidates[i]);}//System.out.println(newcandidates);if(newcandidates.size() == 0)    return res;sum = new ArrayList<Integer>();dfs(target, 0, 0);sum = null;newcandidates = null;return res;    }    public void dfs(int target, int step, int tmpsum){    if(step == newcandidates.size())    {        if(tmpsum == target)    {    if(!res.contains(sum))    res.add(new ArrayList<Integer>(sum));    }    return ;    }if(tmpsum > target)return;sum.add(newcandidates.get(step));dfs(target, step + 1, tmpsum + newcandidates.get(step));sum.remove(sum.size() - 1);dfs(target, step + 1, tmpsum);return;}}

以下代码不超时：

public class Solution {    public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();if(candidates.length == 0)return res;Arrays.sort(candidates);ArrayList<Integer> sum = new ArrayList<Integer>();dfs(res, candidates, target, sum, 0, 0);return res;    }    public void dfs(ArrayList<ArrayList<Integer>> res, int[] array, int target, ArrayList<Integer> sum, int step, int tmpsum){if(tmpsum == target){if(!res.contains(sum))res.add(new ArrayList<Integer>(sum));return;}if(tmpsum > target)return;for(int i = step; i < array.length; i++){sum.add(array[i]);dfs(res, array, target, sum, i, tmpsum + array[i]);sum.remove(sum.size() - 1);}return;}}

如果循环里面的 dfs 函数 里，循环步长 i 变成 i +1 ，那么就变形为每个数只能用一次。