《leetCode》：Combination Sum

题目

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] 

题目大意：在一个全部是正数的数组找出所有等于target的组合，其中每个数可以用无限次。

实现代码如下：

/*思想：先对数组进行排序，然后进行递归，递归的思想为：if(candidates[i]<target){//将candidates[i]加入到所有其满足combination(candidates,target-candidates[i],i)的List中                for(List<Integer> l:combination(candidates,target-candidates[i],i)){                    l.add(0, candidates[i]);                    result.add(l);                }            }*/public class Solution {    public List<List<Integer>>  combination(int []candidates,int target,int start){        if(candidates==null||candidates.length<1){            return null;        }        List<List<Integer>> result=new ArrayList<List<Integer>>();        //从start位置开始寻找，找出所有满足target的所有情况        for(int i=start;i<candidates.length;i++){            if(candidates[i]<target){//将candidates[i]加入到所有其满足combination(candidates,target-candidates[i],i)的List中                for(List<Integer> l:combination(candidates,target-candidates[i],i)){                    l.add(0, candidates[i]);                    result.add(l);                }            }            else if(candidates[i]==target){                List<Integer> tempList=new ArrayList();                tempList.add(candidates[i]);                result.add(tempList);                return result;//直接返回就可以了，当然也可以不加这条语句            }            else{                break;//如果当前值大于target时，后面的值一定也大于target，因此，退出循环即可。            }        }        return result;    }    public List<List<Integer>> combinationSum(int[] candidates, int target) {        if(candidates==null||candidates.length<1){            return null;        }        //第一步：对数组进行排序        Arrays.sort(candidates);        return combination(candidates,target,0);    }}

AC结果如下