leetcode 39 Combination Sum (回溯法)

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

Given a set of candidate numbers (C) (without duplicates) 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:

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

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这道题采用回溯法的思想。重点在于，此题中要求结果不能重复，比如有了[1,2,3]那就不能有[2,1,3]。只要数字一样就是一样的。这样在递归时候，需要让新的索引始终在之前索引之后，这样就能避免因为新索引在原先索引之前造成重复。

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

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